itertools模块的使用

#无限迭代器count
from itertools import *
a=count()
results=[]
while True:
    val=a.__next__()
    results.append(val)
    if len(results)==10:
        break
print(results)

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

#repeat
res=repeat('a',10)
for i in res:
    print(i)

a
a
a
a
a
a
a
a
a
a

#cycle
cycle('abc')

<itertools.cycle at 0x5381a68>

res=accumulate([1,2,3])
for i in res:
    print(i)

1
3
6

#chain
l1=[1,2,3]
l2=[4,5,6]
for i in chain(l1,l2):
    print(i)

1
2
3
4
5
6

for i in dropwhile(lambda x: x<5, [1,4,6,4,1]):
    print(i)  #前面开始的小于5的数被丢弃，后面的不计算

6
4
1

#takewhile保留符合条件的
for i in takewhile(lambda x: x<5, [1,4,6,4,1]):
    print(i)

1
4

#切片
for i in islice('ABCDEFG', 2,4):
    print(i)

C
D

for i in compress('ABCDEF', [1,None,1,0,1,1]):
    print(i)

A
C
E
F

#zip_longest(),类似于zip,多余的项为none或指定
for i,j in dict(zip_longest('ABCD', 'xy',fillvalue='novalue')).items():
    print(i,j)

A x
B y
C novalue
D novalue

#tee()，返回多个独立迭代器
i1,i2=tee(range(5))
print(list(i1))
print(list(i2))

[0, 1, 2, 3, 4]
[0, 1, 2, 3, 4]

#groupby,分组
#返回姓名开头的分组数据
from operator import itemgetter
def get_startname(name):
    if name.split()[0]=='zhang':
        return 'zh'
    elif name.split()[0]=='li':
        return 'li'

d1={"id":"001",
    "name":"zhang san"
}
d2={"id":"002",
    "name":"li si"
}
d3={"id":"003",
    "name":"zhang wu"
}
l1=[]
l1.append(d1)
l1.append(d2)
l1.append(d3)
l1 = sorted(l1,key = itemgetter('name'))
datas=groupby(l1, lambda p:get_startname(p['name']))
for key,group in datas:
    print(key,list(group))

li [{'id': '002', 'name': 'li si'}]
zh [{'id': '001', 'name': 'zhang san'}, {'id': '003', 'name': 'zhang wu'}]