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pandas_sum_tricks.ipynb
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450
pandas_sum_tricks.ipynb
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{
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"metadata": {
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"name": "",
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"signature": "sha256:8222de4af96dc6569eddec8d75df6855e8bac273e12e8739fffc42aafd712ba2"
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"input": [
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"%load_ext watermark \n",
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"%watermark -d -v -a 'Sebastian Raschka' -p numpy,pandas"
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],
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"language": "python",
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"metadata": {},
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"outputs": [
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{
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"output_type": "stream",
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"stream": "stdout",
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"text": [
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"Sebastian Raschka 23/12/2014 \n",
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"\n",
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"CPython 3.4.2\n",
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"IPython 2.3.1\n",
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"\n",
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"numpy 1.9.1\n",
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"pandas 0.15.2\n"
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]
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}
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],
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"prompt_number": 1
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"<br>\n",
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"<br>"
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]
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},
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{
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"cell_type": "heading",
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"level": 1,
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"metadata": {},
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"source": [
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"4 Simple Tricks To Speed up the Sum Calculation in Pandas"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"I wanted to improve the performance of some passages in my code a little bit and found that some simple tweaks can speed up the `pandas` section significantly. I thought that it might be one useful thing to share -- and no Cython or just-in-time compilation is required! "
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"<br>\n",
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"<br>"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"In my case, I had a large dataframe where I wanted to calculate the sum of specific columns for different combinations of rows (approx. 100,000,000 of them, that's why I was looking for ways to speed it up). Anyway, below is a simple toy DataFrame to explore the `.sum()` method a little bit."
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]
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},
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{
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"cell_type": "code",
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"collapsed": false,
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"input": [
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"import pandas as pd\n",
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"import numpy as np\n",
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"\n",
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"df = pd.DataFrame()\n",
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"\n",
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"for col in ('a', 'b', 'c', 'd'):\n",
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" df[col] = pd.Series(range(1000), index=range(1000))"
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],
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"language": "python",
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"metadata": {},
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"outputs": [],
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"prompt_number": 2
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},
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{
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"cell_type": "code",
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"collapsed": false,
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"input": [
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"df.tail()"
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],
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"language": "python",
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"metadata": {},
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"outputs": [
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{
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"html": [
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"<div style=\"max-height:1000px;max-width:1500px;overflow:auto;\">\n",
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"<table border=\"1\" class=\"dataframe\">\n",
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" <thead>\n",
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" <tr style=\"text-align: right;\">\n",
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" <th></th>\n",
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" <th>a</th>\n",
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" <th>b</th>\n",
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" <th>c</th>\n",
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" <th>d</th>\n",
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" </tr>\n",
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" </thead>\n",
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" <tbody>\n",
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" <tr>\n",
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" <th>995</th>\n",
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" <td> 995</td>\n",
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" <td> 995</td>\n",
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" <td> 995</td>\n",
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" <td> 995</td>\n",
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" </tr>\n",
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" <tr>\n",
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" <th>996</th>\n",
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" <td> 996</td>\n",
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" <td> 996</td>\n",
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" <td> 996</td>\n",
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" <td> 996</td>\n",
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" </tr>\n",
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" <tr>\n",
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" <th>997</th>\n",
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" <td> 997</td>\n",
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" <td> 997</td>\n",
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" <td> 997</td>\n",
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" <td> 997</td>\n",
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" </tr>\n",
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" <tr>\n",
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" <th>998</th>\n",
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" <td> 998</td>\n",
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" <td> 998</td>\n",
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" <td> 998</td>\n",
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" <td> 998</td>\n",
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" </tr>\n",
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" <tr>\n",
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" <th>999</th>\n",
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" <td> 999</td>\n",
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" <td> 999</td>\n",
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" <td> 999</td>\n",
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" <td> 999</td>\n",
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" </tr>\n",
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" </tbody>\n",
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"</table>\n",
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"</div>"
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],
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"metadata": {},
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"output_type": "pyout",
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"prompt_number": 3,
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"text": [
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" a b c d\n",
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"995 995 995 995 995\n",
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"996 996 996 996 996\n",
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"997 997 997 997 997\n",
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"998 998 998 998 998\n",
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"999 999 999 999 999"
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]
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}
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],
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"prompt_number": 3
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"<br>"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Let's assume we are interested in calculating the sum of column `a`, `c`, and `d`, which would look like this:"
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]
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},
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{
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"cell_type": "code",
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"collapsed": false,
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"input": [
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"df.loc[:, ['a', 'c', 'd']].sum(axis=0)"
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],
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"language": "python",
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"metadata": {},
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"outputs": [
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{
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"metadata": {},
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"output_type": "pyout",
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"prompt_number": 4,
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"text": [
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"a 499500\n",
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"c 499500\n",
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"d 499500\n",
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"dtype: int64"
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]
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}
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],
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"prompt_number": 4
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Now, the `.loc` method is probably the most \"costliest\" one for this kind of operation. Since we are only intersted in the resulting numbers (i.e., the column sums), there is no need to make a copy of the array. Anyway, let's use the method above as a reference for comparison:"
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]
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},
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{
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"cell_type": "code",
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"collapsed": false,
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"input": [
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"# 1\n",
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"%timeit -n 1000 -r 5 df.loc[:, ['a', 'c', 'd']].sum(axis=0)"
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],
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"language": "python",
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"metadata": {},
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"outputs": [
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{
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"output_type": "stream",
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"stream": "stdout",
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"text": [
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"1000 loops, best of 5: 1.28 ms per loop\n"
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]
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}
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],
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"prompt_number": 5
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"<br>"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Although this is a rather small DataFrame (1000 x 4), let's see by how much we can speed it up using a different slicing method:"
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]
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},
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{
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"cell_type": "code",
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"collapsed": false,
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"input": [
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"# 2\n",
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"%timeit -n 1000 -r 5 df[['a', 'c', 'd']].sum(axis=0)"
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],
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"language": "python",
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"metadata": {},
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"outputs": [
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{
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"output_type": "stream",
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"stream": "stdout",
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"text": [
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"1000 loops, best of 5: 1.03 ms per loop\n"
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]
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}
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],
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"prompt_number": 6
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"<br>"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Next, let us use the Numpy representation of the `NDFrame` via the `.values` attribue:"
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]
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},
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{
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"cell_type": "code",
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"collapsed": false,
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"input": [
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"# 3\n",
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"%timeit -n 1000 -r 5 df[['a', 'c', 'd']].values.sum(axis=0)"
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],
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"language": "python",
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"metadata": {},
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"outputs": [
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{
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"output_type": "stream",
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"stream": "stdout",
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"text": [
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"1000 loops, best of 5: 721 \u00b5s per loop\n"
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]
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}
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],
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"prompt_number": 7
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"<br>"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"While the speed improvements in #2 and #3 were not really a surprise, the next \"trick\" surprised me a little bit. Here, we are calculating the sum of each column separately rather than slicing the array."
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]
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},
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{
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"cell_type": "code",
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"collapsed": false,
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"input": [
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"[df[col].values.sum(axis=0) for col in ('a', 'c', 'd')]"
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],
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"language": "python",
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"metadata": {},
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"outputs": [
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{
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"metadata": {},
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"output_type": "pyout",
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"prompt_number": 8,
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"text": [
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"[499500, 499500, 499500]"
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]
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}
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],
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"prompt_number": 8
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},
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{
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"cell_type": "code",
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"collapsed": false,
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"input": [
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"# 4\n",
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"%timeit -n 1000 -r 5 [df[col].values.sum(axis=0) for col in ('a', 'c', 'd')]"
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],
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"language": "python",
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"metadata": {},
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"outputs": [
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{
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"output_type": "stream",
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"stream": "stdout",
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"text": [
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"1000 loops, best of 5: 64.8 \u00b5s per loop\n"
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]
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}
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],
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"prompt_number": 9
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"In this case, this is an almost 10x improvement!"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"<br>"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"One more thing: Let's try the Einstein summation convention [`einsum`](http://docs.scipy.org/doc/numpy/reference/generated/numpy.einsum.html)."
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]
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},
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{
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"cell_type": "code",
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"collapsed": false,
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"input": [
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"from numpy import einsum\n",
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"[einsum('i->', df[col].values) for col in ('a', 'c', 'd')]"
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],
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"language": "python",
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"metadata": {},
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"outputs": [
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{
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"metadata": {},
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"output_type": "pyout",
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"prompt_number": 10,
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"text": [
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"[499500, 499500, 499500]"
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]
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}
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],
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"prompt_number": 10
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},
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{
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"cell_type": "code",
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"collapsed": false,
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"input": [
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"# 5\n",
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"%timeit -n 1000 -r 5 [einsum('i->', df[col].values) for col in ('a', 'c', 'd')]"
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],
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"language": "python",
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"metadata": {},
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"outputs": [
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{
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"output_type": "stream",
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"stream": "stdout",
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"text": [
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"1000 loops, best of 5: 57.2 \u00b5s per loop\n"
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]
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}
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],
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"prompt_number": 11
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"<br>"
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]
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},
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{
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"cell_type": "heading",
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"level": 3,
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"metadata": {},
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"source": [
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"Conclusion:"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Using some simple tricks, the column sum calculation improved from 1280 to 57.2 \u00b5s per loop (approx. 22x faster!)"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"<br>"
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]
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}
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],
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"metadata": {}
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}
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]
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}
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