#sorted(list) function: which takes a list and returns a new list with those elements in sorted order a = [5, 1, 4, 3] print sorted(a) ## [1, 3, 4, 5] print a ## [5, 1, 4, 3] #The sorted() function can be customized though optional arguments. The sorted() optional argument reverse=True, e.g. sorted(list, reverse=True), makes it sort backwards. strs = [‘aa‘, ‘BB‘, ‘zz‘, ‘CC‘] print sorted(strs) ## [‘BB‘, ‘CC‘, ‘aa‘, ‘zz‘] (case sensitive) print sorted(strs, reverse=True) ## [‘zz‘, ‘aa‘, ‘CC‘, ‘BB‘]2. 使用key=func 进行定制排序
strs = [‘ccc‘, ‘aaaa‘, ‘d‘, ‘bb‘] print sorted(strs, key=len) ## [‘d‘, ‘bb‘, ‘ccc‘, ‘aaaa‘] #2.As another example, specifying "str.lower" as the key function is a way to force the sorting to treat uppercase and lowercase the same: ## "key" argument specifying str.lower function to use for sorting print sorted(strs, key=str.lower) ## [‘aa‘, ‘BB‘, ‘CC‘, ‘zz‘] #3.You can also pass in your own MyFn as the key function, like this: ## Say we have a list of strings we want to sort by the last letter of the string. strs = [‘xc‘, ‘zb‘, ‘yd‘ ,‘wa‘] ## Write a little function that takes a string, and returns its last letter. ## This will be the key function (takes in 1 value, returns 1 value). def MyFn(s): return s[-1] ## Now pass key=MyFn to sorted() to sort by the last letter: print sorted(strs, key=MyFn) ## [‘wa‘, ‘zb‘, ‘xc‘, ‘yd‘]
3. sort()方法
#the sort() method on a list sorts that list into ascending order list.sort() #the sort() method changes the underlying list and returns None
元组tuple
tuple是不可变的，并且大小不能改变
##Tuples play a sort of "struct" role in Python -- a convenient way to pass around a little logical, fixed size bundle of values.
tuple = (1, 2, ‘hi‘) print len(tuple) ## 3 print tuple[2] ## hi tuple[2] = ‘bye‘ ## NO, tuples cannot be changed tuple = (1, 2, ‘bye‘) ## this works, it created a new tuple
nums = [1, 2, 3, 4] squares = [ n * n for n in nums ] ## [1, 4, 9, 16] [ expr for var in nums] -- ## the expr to its left is evaluated once for each element to give the values for the new list.
strs = [‘hello‘, ‘and‘, ‘goodbye‘] shouting = [ s.upper() + ‘!!!‘ for s in strs ] ## [‘HELLO!!!‘, ‘AND!!!‘, ‘GOODBYE!!!‘]
## Select values <= 2 nums = [2, 8, 1, 6] small = [ n for n in nums if n <= 2 ] ## [2, 1] ## Select fruits containing ‘a‘, change to upper case fruits = [‘apple‘, ‘cherry‘, ‘bannana‘, ‘lemon‘] afruits = [ s.upper() for s in fruits if ‘a‘ in s ] ## [‘APPLE‘, ‘BANNANA‘]
##The contents of a dict can be written as a series of key:value pairs within braces { }, e.g. dict = {key1:value1, key2:value2, ... }. The "empty dict" is just an empty pair of curly braces
{}.
字典是python中唯一的映射类型（哈希表）,常用成员方法：
keys()
values()
items()
dic = {"name":"well","age":20,"gender":"male"}
dic.keys() #返回所有的key
dic[‘name‘] #返回 "well" ## Can build up a dict by starting with the the empty dict {}
## and storing key/value pairs into the dict like this:
## dict[key] = value-for-that-key
dict = {}
dict[‘a‘] = ‘alpha‘
dict[‘g‘] = ‘gamma‘
dict[‘o‘] = ‘omega‘
print dict ## {‘a‘: ‘alpha‘, ‘o‘: ‘omega‘, ‘g‘: ‘gamma‘}
print dict[‘a‘] ## Simple lookup, returns ‘alpha‘
dict[‘a‘] = 6 ## Put new key/value into dict
‘a‘ in dict ## True
## print dict[‘z‘] ## Throws KeyError
if ‘z‘ in dict: print dict[‘z‘] ## Avoid KeyError
print dict.get(‘z‘) ## None (instead of KeyError)
## By default, iterating over a dict iterates over its keys.
## Note that the keys are in a random order.
for key in dict: print key
## prints a g o
## Exactly the same as above
for key in dict.keys(): print key
## Get the .keys() list:
print dict.keys() ## [‘a‘, ‘o‘, ‘g‘]
## Likewise, there‘s a .values() list of values
print dict.values() ## [‘alpha‘, ‘omega‘, ‘gamma‘]
## Common case -- loop over the keys in sorted order,
## accessing each key/value
for key in sorted(dict.keys()):
print key, dict[key]
## .items() is the dict expressed as (key, value) tuples
print dict.items() ## [(‘a‘, ‘alpha‘), (‘o‘, ‘omega‘), (‘g‘, ‘gamma‘)]
## This loop syntax accesses the whole dict by looping
## over the .items() tuple list, accessing one (key, value)
## pair on each iteration.
for k, v in dict.items(): print k, ‘>‘, v
## a > alpha o > omega g > gamma hash = {}
hash[‘word‘] = ‘garfield‘
hash[‘count‘] = 42
s = ‘I want %(count)d copies of %(word)s‘ % hash # %d for int, %s for string
print s
# ‘I want 42 copies of garfield‘Del：
##The "del" operator does deletions. In the simplest case, it can remove the definition of a variable, as if that variable had not been defined. Del can also be used on list elements or slices to delete that part of the
list and to delete entries from a dictionary.
var = 6
del var # var no more!
list = [‘a‘, ‘b‘, ‘c‘, ‘d‘]
del list[0] ## Delete first element
del list[-2:] ## Delete last two elements
print list ## [‘b‘]
dict = {‘a‘:1, ‘b‘:2, ‘c‘:3}
del dict[‘b‘] ## Delete ‘b‘ entry
print dict ## {‘a‘:1, ‘c‘:3}# Echo the contents of a file f = open(‘foo.txt‘, ‘rU‘) ## The special mode ‘rU‘ is the "Universal" option for text files where it‘s smart about converting different line-endings so they always come through as a simple ‘\n‘. ## Instead of ‘r‘, use ‘w‘ for writing, and ‘a‘ for append. for line in f: ## iterates over the lines of the file print line, ## trailing, so print does not add an end-of-line char ## since ‘line‘ already includes the end-of line. f.close()## The f.readlines() method reads the whole file into memory and returns its contents as a list of its lines.
读取unicode文件，可以使用 codecs 模块：
import codecs f = codecs.open(‘foo.txt‘, ‘rU‘, ‘utf-8‘) for line in f: # here line is a *unicode* string
word1 count1
word2 count2
......
def word_count_dict(filename):
"""Returns a word/count dict for this filename."""
# Utility used by count() and Topcount().
word_count = {} # Map each word to its count
input_file = open(filename, ‘r‘)
for line in input_file:
words = line.split()
for word in words:
word = word.lower()
# Special case if we‘re seeing this word for the first time.
if not word in word_count:
word_count[word] = 1
else:
word_count[word] = word_count[word] + 1
input_file.close() # Not strictly required, but good form.
return word_count原文：http://blog.csdn.net/tao_sun/article/details/18744081