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added cartesian product
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3 files changed

+299
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Diff for: ‎.ipynb_checkpoints/cartesian_product-checkpoint.ipynb

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{
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"metadata": {
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"name": "",
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"signature": "sha256:ade2566ab5c7ca9cb880a39b507a20f21513d98bcb1fdd1be99ec7fe882c3646"
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},
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"nbformat": 3,
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"nbformat_minor": 0,
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"worksheets": [
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{
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"cells": [
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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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"# Cartesian Product\n",
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"\n",
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"This snippit was written by [Chris R. Albon](http://www.chrisralbon.com/) and is part of his collection of [well-documented Python snippits](https://github.com/chrisalbon/code_py). All code is written in Python 3 in iPython notebook and offered under the [Creative Commons Attribution-ShareAlike 4.0 International License](http://creativecommons.org/licenses/by-sa/4.0/\"><img alt=\"Creative Commons License)."
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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 pandas as pd"
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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": 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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"# Create two lists\n",
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"i = [1,2,3,4,5]\n",
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"j = [1,2,3,4,5]"
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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": 11
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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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"# List every single x in i with every single y (i.e. Cartesian product)\n",
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"[(x, y) for x in i for y in j]"
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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": 14,
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"text": [
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"[(1, 1),\n",
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" (1, 2),\n",
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" (1, 3),\n",
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" (1, 4),\n",
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" (1, 5),\n",
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" (2, 1),\n",
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" (2, 2),\n",
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" (2, 3),\n",
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" (2, 4),\n",
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" (2, 5),\n",
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" (3, 1),\n",
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" (3, 2),\n",
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" (3, 3),\n",
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" (3, 4),\n",
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" (3, 5),\n",
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" (4, 1),\n",
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" (4, 2),\n",
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" (4, 3),\n",
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" (4, 4),\n",
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" (4, 5),\n",
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" (5, 1),\n",
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" (5, 2),\n",
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" (5, 3),\n",
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" (5, 4),\n",
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" (5, 5)]"
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]
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}
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],
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"prompt_number": 14
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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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"# An alternative way to do the cartesian product\n",
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"\n",
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"# import itertools\n",
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"import itertools\n",
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"\n",
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"# for two sets, find the the cartisan product\n",
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"for i in itertools.product([1,2,3,4,5], [1,2,3,4,5]):\n",
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" # and print it\n",
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" print(i)"
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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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"(1, 1)\n",
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"(1, 2)\n",
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"(1, 3)\n",
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"(1, 4)\n",
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"(1, 5)\n",
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"(2, 1)\n",
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"(2, 2)\n",
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"(2, 3)\n",
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"(2, 4)\n",
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"(2, 5)\n",
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"(3, 1)\n",
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"(3, 2)\n",
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"(3, 3)\n",
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"(3, 4)\n",
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"(3, 5)\n",
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"(4, 1)\n",
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"(4, 2)\n",
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"(4, 3)\n",
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"(4, 4)\n",
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"(4, 5)\n",
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"(5, 1)\n",
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"(5, 2)\n",
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"(5, 3)\n",
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"(5, 4)\n",
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"(5, 5)\n"
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]
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}
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],
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"prompt_number": 16
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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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}

Diff for: ‎Untitled0.ipynb

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{
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"metadata": {
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"name": "",
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"signature": "sha256:24d12ab64e5e93c82e0acf22e043d94d7006b033acc4160a5a99e702b6873929"
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},
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"nbformat": 3,
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"nbformat_minor": 0,
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"worksheets": []
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}

Diff for: ‎cartesian_product.ipynb

+145
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,145 @@
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{
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"metadata": {
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"name": "",
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"signature": "sha256:ade2566ab5c7ca9cb880a39b507a20f21513d98bcb1fdd1be99ec7fe882c3646"
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},
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"nbformat": 3,
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"nbformat_minor": 0,
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"worksheets": [
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{
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"cells": [
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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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"# Cartesian Product\n",
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"\n",
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"This snippit was written by [Chris R. Albon](http://www.chrisralbon.com/) and is part of his collection of [well-documented Python snippits](https://github.com/chrisalbon/code_py). All code is written in Python 3 in iPython notebook and offered under the [Creative Commons Attribution-ShareAlike 4.0 International License](http://creativecommons.org/licenses/by-sa/4.0/\"><img alt=\"Creative Commons License)."
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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 pandas as pd"
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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": 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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"# Create two lists\n",
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"i = [1,2,3,4,5]\n",
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"j = [1,2,3,4,5]"
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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": 11
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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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"# List every single x in i with every single y (i.e. Cartesian product)\n",
50+
"[(x, y) for x in i for y in j]"
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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": 14,
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"text": [
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"[(1, 1),\n",
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" (1, 2),\n",
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" (1, 3),\n",
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" (1, 4),\n",
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" (1, 5),\n",
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" (2, 1),\n",
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" (2, 2),\n",
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" (2, 3),\n",
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" (2, 4),\n",
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" (2, 5),\n",
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" (3, 1),\n",
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" (3, 2),\n",
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" (3, 3),\n",
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" (3, 4),\n",
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" (3, 5),\n",
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" (4, 1),\n",
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" (4, 2),\n",
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" (4, 3),\n",
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" (4, 4),\n",
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" (4, 5),\n",
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" (5, 1),\n",
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" (5, 2),\n",
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" (5, 3),\n",
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" (5, 4),\n",
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" (5, 5)]"
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]
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}
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],
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"prompt_number": 14
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},
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{
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"cell_type": "code",
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"collapsed": false,
93+
"input": [
94+
"# An alternative way to do the cartesian product\n",
95+
"\n",
96+
"# import itertools\n",
97+
"import itertools\n",
98+
"\n",
99+
"# for two sets, find the the cartisan product\n",
100+
"for i in itertools.product([1,2,3,4,5], [1,2,3,4,5]):\n",
101+
" # and print it\n",
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" print(i)"
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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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"(1, 1)\n",
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"(1, 2)\n",
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"(1, 3)\n",
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"(1, 4)\n",
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"(1, 5)\n",
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"(2, 1)\n",
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"(2, 2)\n",
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"(2, 3)\n",
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"(2, 4)\n",
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"(2, 5)\n",
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"(3, 1)\n",
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"(3, 2)\n",
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"(3, 3)\n",
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"(3, 4)\n",
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"(3, 5)\n",
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"(4, 1)\n",
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"(4, 2)\n",
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"(4, 3)\n",
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"(4, 4)\n",
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"(4, 5)\n",
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"(5, 1)\n",
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"(5, 2)\n",
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"(5, 3)\n",
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"(5, 4)\n",
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"(5, 5)\n"
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]
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}
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],
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"prompt_number": 16
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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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