简单介绍Python中的配对函数zip()

新建成一个元组构成的列表，它生成列表长度由最短的序列决定：

#zip
seq1=['foo','bar','baz']
seq2=['one','two','three']
zipped=zip(seq1,seq2)
list(zipped)
>>>
>[('foo', 'one'), ('bar', 'two'), ('baz', 'three')]
seq3=[True,False]
zipped2=zip(seq1,seq2,seq3)
list(zipped2)
>>>
>
[('foo', 'one', True), ('bar', 'two', False)]

有时和enumerate搭配使用：

mapping={}
for i,(a,b) in enumerate(zip(seq1,seq2)):
    print('{0},{1},{2}'.format(i,a,b))
    mapping[i]=(a,b)
print(mapping)
>>>
>0,foo,one
1,bar,two
2,baz,three
{0: ('foo', 'one'), 1: ('bar', 'two'), 2: ('baz', 'three')}

利用zip函数进行拆分

pitchers=[('a','1'),('b','2'),('c','3')]
first_names,last_names=zip(*pitchers)
print(first_names,last_names)
>>>
>('a', 'b', 'c') ('1', '2', '3')

zip函数主要用来将n个list中相同location的元素取出来，组成新的data

years = [22,23,24]
names = ['hjw','hyr','lyq']
a = zip(names,years) 
print(a)        # 
print(list(a))  # [('hjw', 22), ('hyr', 23), ('lyq', 24)]
data = []
for (year,name) in zip(years,names):
    data.append({'name':name,'year':year})
print(data)
# [{'name': 'hjw', 'year': 22}, {'name': 'hyr', 'year': 23}, {'name': 'lyq', 'year': 24}]