Chapter 9 Dictionaries

List: A linear collection of values that stay in order

       Lists index their entries based on the position in the list

Dictionary: A "bag" of values, each with its own label

       Dictionaries are like bags - no order

       So we index the things we put in the dictionary with a "lookup tag"

Dictionaries are like Lists except that theyuse keys instead of numbersto look up values.

purse = dict()  # kind of like, use the label to replace ordered number to mark the placepurse['money'] = 12purse['candy'] =3purse['tissues'] = 75PRint purseprint purse['candy']purse['candy'] = purse['candy'] + 2print purSEOutput:
{'money': 12, 'tissues': 75, 'candy': 3}3{'money': 12, 'tissues': 75, 'candy': 5}
# the label is not always 'string'
Empty dictionary
ooo = {}print ooo
Tracebacks: an error will appear if you refer a key which is not in the dictionary
We can use the in Operator to check if a key is in the dic
print 'csev' in cccoutput:FalseAn interesting example: count names
namelist = ['Joe', 'Joey', 'Joe', 'Mike', 'Mike', 'Joey', 'Mike', 'Yifan', 'Joe']counts = dict()for name in namelist:    if name in counts:        counts[name] = counts[name] + 1    else:        counts[name] = 1print countsAnother way might be more understandable:
for name in namelist:    if name not in counts:        counts[name] = 1    else:        counts[name] = counts[name] + 1output:
{'Yifan': 1, 'Mike': 3, 'Joe': 3, 'Joey': 2}
The get method for dictionary
If the key is not in the dictionary, make the value equal to aDefault Value
The following two code snippets have the same result
if name in counts:    print counts[name]else:    print 0
print count.get(name,0)Simplified counting with get()
namelist = ['Joe', 'Joey', 'Joe', 'Mike', 'Mike', 'Joey', 'Mike', 'Yifan', 'Joe']counts = dict()for name in namelist:    counts[name] = counts.get(name,0) + 1print counts
Def Loops and Dictionaries
for loop cango through all of the keys in the dic and look up the values by keys
namecount = {'Yifan': 1, 'Mike': 3, 'Joe': 3, 'Joey': 2}for key in namecount:    print key, namecount[key]output:
Yifan 1Mike 3Joe 3Joey 2
Get the list of keys (or values, or both)
namecount = {'Yifan': 1, 'Mike': 3, 'Joe': 3, 'Joey': 2}print list(namecount)print namecount.keys()print namecount.values()print namecount.items()output:
['Yifan', 'Mike', 'Joe', 'Joey']['Yifan', 'Mike', 'Joe', 'Joey'][1, 3, 3, 2]         # same order with the keys[('Yifan', 1), ('Mike', 3), ('Joe', 3), ('Joey', 2)]       <- ( , ) is a kind of tuples
Two Iteration Variables!
key-value in pairs
namecount = {'Yifan': 1, 'Mike': 3, 'Joe': 3, 'Joey': 2}for name,count in namecount.items():    print name, count
output:
Yifan 1Mike 3Joe 3Joey 2
An Exercise: Get the most frequently appeared Word in the text and count the number of it
file_name = raw_input("Enter the file name:")fhand = open(file_name,'r')content = fhand.read()wordslist = content.split()count = dict()for word in wordslist:    count[word] = count.get(word,0) + 1biggestcount = Nonebiggestword = Nonefor word, wordcount in count.items():    if biggestword == None or wordcount > biggestcount :        biggestcount = wordcount        biggestword = wordprint biggestword, biggestcountComment: Do not forget the split() function.
Chapter 10 Tuples
Tuples are another kind of sequence that function much like a list
- they have elements which are indexed starting at 0
list: [  ]
tupes: ( )
But,
Tuples are none-changeable, immutable. Kind of similar to a string.
z = (5, 4, 3)         z[2] = 0.      z.sort()
Tuples are more efficient
(x, y) = (100, 4)print y
Conduct the assignment at a time
Tuples are Comparable
print (0, 1, 2) < (5, 1, 2)print (0, 1, 2) < (0, 3, 10000)print ('Jones', 'Sally') < ('Jones', 'Fred')print ('Jones', 'Sally') > ('Jane', 'Sally')print ('Jones', 'Sally') > ('Adam', 'Sally')output:
TrueTrueFalseTrueTrue
Sorting Lists of Tuples
We can take advantage of the ability to sort a list of tuples toget a sorted version of a dictionary
#use d.items() and t.sort()namedic = {'Yifan': 1, 'Mike': 3, 'Joe': 3, 'Joey': 2}nametuple = namedic.items()print nametuple    # sort by keysnametuple.sort()print nametuple#use d.items() and sorted(). More directlynamedic = {'Yifan': 1, 'Mike': 3, 'Joe': 3, 'Joey': 2}print namedic.items()t = sorted(namedic.items())print tfor k,v in sorted(namedic.items()):    print k,voutput:
[('Yifan', 1), ('Mike', 3), ('Joe', 3), ('Joey', 2)][('Joe', 3), ('Joey', 2), ('Mike', 3), ('Yifan', 1)][('Yifan', 1), ('Mike', 3), ('Joe', 3), ('Joey', 2)][('Joe', 3), ('Joey', 2), ('Mike', 3), ('Yifan', 1)]Joe 3Joey 2Mike 3Yifan 1
Sort by values instead of key
namedic = {'Yifan': 1, 'Mike': 3, 'Joe': 3, 'Joey': 2}tmp = list()for k, v in namedic.items():    tmp.append((v,k))print tmptmp.sort(reverse=True)  #(reverse=True) means it presents from the highest to lowestprint tmpoutput:
[(1, 'Yifan'), (3, 'Mike'), (3, 'Joe'), (2, 'Joey')][(3, 'Mike'), (3, 'Joe'), (2, 'Joey'), (1, 'Yifan')]
it seems that the tuple sort focus on the first stuff in each item: (focus, xxx)
An Exercise: The top 10 most common words
file_name = raw_input("Enter the file name:")fhand = open(file_name,'r')content = fhand.read()wordslist = content.split()count = dict()for word in wordslist:    count[word] = count.get(word,0) + 1lll = list()for k,v in count.items():    lll.append((v,k))lll.sort(reverse=True)for i in range(10):    print lll[i]output:
(352, 'Jan')(324, '2008')(245, 'by')(243, 'Received:')(219, '-0500')(218, 'from')(203, '4')(194, 'with')(183, 'Fri,')(136, 'id')
for v,k in lll[:10]:    print v,koutput:
352 Jan324 2008245 by243 Received:219 -0500218 from203 4194 with183 Fri,136 id
Even Shorter Version
namedic = {'Yifan': 1, 'Mike': 3, 'Joe': 3, 'Joey': 2}print sorted([(v,k) for k,v in namedic.items()]) #here '[]' is a list comprehensionoutput:
[(1, 'Yifan'), (2, 'Joey'), (3, 'Joe'), (3, 'Mike')]
Comments:
    List comprehension: creates a dynamic list that meets certain requirements.
              In this case, we make a list of reversed tuples and then sort it.