# zip 案例l1 = [1,2,3,4,5]l2 = [11,22,33,44,55]
z = zip(l1, l2)
print(type(z))
for i in z:    print(i)

<class 'zip'>(1, 11)(2, 22)(3, 33)(4, 44)(5, 55)

l1 = ["wangwang", "mingyue", "yyt"]l2 = [89, 23, 78]
z = zip(l1, l2)
for i in z:    print(i)   l3 = (i for i in z)print(l3)

('wangwang', 89)('mingyue', 23)('yyt', 78)<generator object <genexpr> at 0x000001ED9F1F3ED0>

l1 = [11,22,33,44,55]
em = enumerate(l1)
l2 = [i for i in em]print(l2)

[(0, 11), (1, 22), (2, 33), (3, 44), (4, 55)]

em = enumerate(l1, start=100)
l2 = [i for i in em]print(l2)

[(100, 11), (101, 22), (102, 33), (103, 44), (104, 55)]

import collectionsPoint = collections.namedtuple("Point", ['x', 'y'])p = Point(11, 22)print(p.x)print(p[0])

1111

Circle = collections.namedtuple("Circle", ['x', 'y', 'r'])
c = Circle(100, 120, 20)print(c)print(type(c))
# 检测一下namedtuple到底属于谁的子类isinstance(c, tuple)

Circle(x=100, y=120, r=20)<class '__main__.Circle'>




True

import collectionshelp(collections.namedtuple)

Help on function namedtuple in module collections:
namedtuple(typename, field_names, *, rename=False, defaults=None, module=None)    Returns a new subclass of tuple with named fields.        >>> Point = namedtuple('Point', ['x', 'y'])    >>> Point.__doc__                   # docstring for the new class    'Point(x, y)'    >>> p = Point(11, y=22)             # instantiate with positional args or keywords    >>> p[0] + p[1]                     # indexable like a plain tuple    33    >>> x, y = p                        # unpack like a regular tuple    >>> x, y    (11, 22)    >>> p.x + p.y                       # fields also accessible by name    33    >>> d = p._asdict()                 # convert to a dictionary    >>> d['x']    11    >>> Point(**d)                      # convert from a dictionary    Point(x=11, y=22)    >>> p._replace(x=100)               # _replace() is like str.replace() but targets named fields    Point(x=100, y=22)

from collections import deque
q = deque(['a', 'b', 'c'])print(q)
q.append('d')print(q)
q.appendleft('x')print(q)

deque(['a', 'b', 'c'])deque(['a', 'b', 'c', 'd'])deque(['x', 'a', 'b', 'c', 'd'])

d1 = {"one":1, "two":2, "three":3}print(d1['one'])print(d1['four'])

1


---------------------------------------------------------------------------
KeyError                                  Traceback (most recent call last)
<ipython-input-34-83fcfcbfcf65> in <module>      1 d1 = {"one":1, "two":2, "three":3}      2 print(d1['one'])----> 3 print(d1['four'])

KeyError: 'four'

from collections import defaultdict# lambda表达式，直接返回字符串func = lambda: "ruochen"d2 = defaultdict(func)
d2["one"] = 1d2["two"] = 2
print(d2['one'])print(d2['four'])

1ruochen

from collections import Counter
# 为什么下面结果不把abcd...作为键值，而是以其中每一个字母作为键值# 需要括号里内容为可迭代c = Counter("abcdeabcabc")print(c)

Counter({'a': 3, 'b': 3, 'c': 3, 'd': 1, 'e': 1})

# help(Counter)

s = {"I", "love", "you"}c = Counter(s)
print(c)

Counter({'love': 1, 'you': 1, 'I': 1})