每天十分钟,十二天入门Python(十二)
collections集合模块
from collections import namedtuple
Point = namedtuple('Point',['x','y'])
p = Point(1,2)
print p.x
print p.y
p.z = 3
# 1
# 2
# Traceback (most recent call last):
# File "/Users/apple/workspace/pyDemo/demo.py", line 11, in <module>
# p.z = 3
# AttributeError: 'Point' object has no attribute 'z'
namedtuple()实际上就是创建了一个tuple对象,但是可以根据属性直接引用,属性值和属性个数都不可变,和tuple的不变性一致.
from collections import deque
arr = deque([1,2,3,4,5])
arr.append(6)
arr.appendleft(0)
print arr
# deque([0, 1, 2, 3, 4, 5, 6])
deque()除了实现list的append()和pop()外，还支持appendleft()和popleft()，这样就可以非常高效地往头部添加或删除元素.
from collections import defaultdict
dd = defaultdict(lambda: 'N/A')
dd['key1'] = 'abc'
print dd['key1'] # key1存在
# 'abc'
print dd['key2'] # key2不存在，返回默认值
# 'N/A'
defaultdict()相比于dict的进步就是在获取不存在的key的时候不会报错~
from collections import OrderedDict
d = dict([('name','zhangsan'),('age',22)])
print d
d2 = OrderedDict([('name','zhangsan'),('age',22)])
print d2
# {'age': 22, 'name': 'zhangsan'}
# OrderedDict([('name', 'zhangsan'), ('age', 22)])
o = OrderedDict()
o['x'] = 1
o['z'] = 3
o['y'] = 2
print o.keys()
print o
# ['x', 'z', 'y']
# OrderedDict([('x', 1), ('z', 3), ('y', 2)])
OrderDict()创建了有序的dict,其中key的顺序和插入的顺序一致.
from collections import Counter
c = Counter()
for char in 'hello world':
c[char] = c[char]+1
print c
# Counter({'l': 3, 'o': 2, ' ': 1, 'e': 1, 'd': 1, 'h': 1, 'r': 1, 'w': 1})
Counter()用来计数,比如上面的demo记录了每个字母出现的次数.
itertools迭代模块
import itertools
ns = itertools.count()
for n in ns:
print n
# 1
# 2
# 3
# ...
# count()会创建一个无限迭代器
ns = itertools.cycle([1,2,3])
for n in ns:
print n
# 1
# 2
# 3
# 1
# 2
# 3
# ...
# cycle会无限的重复指定的序列
ns = itertools.repeat([1,2,3],2)
for n in ns:
print n
# [1, 2, 3]
# [1, 2, 3]
# repeat的内容会重复循环,第二个参数可以设定循环次数
for n in itertools.chain('QWE','RTY','U'):
print n
# Q
# W
# E
# R
# T
# Y
# U
# chain()可以将可迭代对象串起来,形成更大的迭代器
for key,group in itertools.groupby('AADDWWCCVVCCEEAA'):
print key,list(group)
print '--------------------------------------------------'
for key2,group2 in itertools.groupby('AADDWWCCVVCCEEAAaaccswww',lambda c:c.upper()):
print key2,list(group2)
# A ['A', 'A']
# D ['D', 'D']
# W ['W', 'W']
# C ['C', 'C']
# V ['V', 'V']
# C ['C', 'C']
# E ['E', 'E']
# A ['A', 'A']
# --------------------------------------------------
# A ['A', 'A']
# D ['D', 'D']
# W ['W', 'W']
# C ['C', 'C']
# V ['V', 'V']
# C ['C', 'C']
# E ['E', 'E']
# A ['A', 'A', 'a', 'a']
# C ['c', 'c']
# S ['s']
# W ['w', 'w', 'w']
# 一般情况下groupby()仅能将相邻重复的元素挑出来,实际上挑选规则是通过函数完成的，只要作用于函数的两个元素返回的值相等，这两个元素就被认为是在一组的，而函数返回值作为组的key。如果我们要忽略大小写分组，就可以让元素'A'和'a'都返回相同的key