{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Dictionaries\n",
    "\n",
    "Python dictionaries are analogous to Java's HashMap data structure, and like HashMap, they provide fast access to a value associated with a key using a hash function.  And this being Python, they're a lot easier to use!"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# first, let's create an empty dictionary\n",
    "eng2sp = {}\n",
    "# and then let's add a few items to it\n",
    "eng2sp['one'] = 'uno'\n",
    "eng2sp['two'] = 'dos'\n",
    "eng2sp['one']"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "eng = ['one', 'two']\n",
    "sp = ['uno', 'dos']\n",
    "sp[0]"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "You can create a dictionary by providing key-value pairs in the same format as the output we saw above:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "eng2sp = {'one': 'uno', 'two': 'dos', 'three': 'tres'}"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Note that order doesn't matter!"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "inventory = {'apples': 430, 'bananas': 312, 'oranges': 525, 'pears': 217}\n",
    "inventory"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The `len` function returns the number of key-value pairs:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "len(inventory)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "We can change the values in a dictionary:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "inventory['pears'] = 0"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "You can determine if a dictionary has an entry with a given key:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "'pears' in inventory, 'bananas' in inventory\n",
    "'kiwi' in inventory"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "You can use integers, strings and tuples as keys:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "pair_dictionary = {}\n",
    "pair_dictionary[('a', 1)] = 1\n",
    "pair_dictionary[('z', 3)] = 5\n",
    "pair_dictionary[(1,1)] = 3\n",
    "pair_dictionary[10] = 5\n",
    "pair_dictionary"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "hash((0,1))\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Keys of a dictionary have to be immutable, therefore we cannot use lists as keys (but tuples are fine)."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "You can iterate over a dictionary using a for loop:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "inventory = {'apples': 430, 'bananas': 312, 'oranges': 525, 'pears': 217}\n",
    "for fruit in inventory :\n",
    "    print (fruit, inventory[fruit])"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Dictionaries are extremely useful for efficiently keeping track of things.  Let's write a function that computes the number of occurrences of each letter in a sentence:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "{'f': 1,\n",
       " 'h': 1,\n",
       " 'i': 1,\n",
       " 'n': 2,\n",
       " 'o': 1,\n",
       " 'p': 1,\n",
       " 's': 1,\n",
       " 't': 1,\n",
       " 'u': 1,\n",
       " 'y': 1}"
      ]
     },
     "execution_count": 2,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "def letter_count(sentence) :\n",
    "    counts = {}\n",
    "    for letter in sentence.lower() :\n",
    "        if letter in \" !.?,\" :\n",
    "            continue   # ignore whitespace and punctuation\n",
    "        if not(letter in counts) :\n",
    "            counts[letter] = 0\n",
    "        counts[letter] += 1\n",
    "    return counts\n",
    "letter_count('Python is fun!')\n",
    " "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "We can modify the above function to use the ``get`` method of a dictionary:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "def letter_count(sentence) :\n",
    "    counts = {}\n",
    "    for letter in sentence.lower() :\n",
    "        if letter in \" !.?,\\n\\t\" :\n",
    "            continue   # ignore whitespace and punctuation\n",
    "        counts[letter] = counts.get(letter, 0) + 1\n",
    "    return counts\n",
    "letter_count('Python is fun!')\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "eng2sp = {'one': 'uno', 'two': 'dos', 'three': 'tres'}\n",
    "eng2sp.get('tres', 0)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The second argument of the ``get`` method of a dictionary is the value which should be used in case the given key is not in the dictionary, which allowed us to skip the step that handles the case of a kmer that is not in the dictionary.\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Exercises:\n",
    "\n",
    "* Most common substring.  Write a function called `most_common_substring(s, length)` that returns the substring of the given length that occurs the most number of times within the input string `s`.  For example, on the input `'mississipi', 4`, the return value should be `'issi'`.  Hint:  use slices to extract substrings of the appropriate length."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": []
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python 3",
   "language": "python",
   "name": "python3"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 3
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython3",
   "version": "3.6.3"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 1
}