Question

请允许我说,我有两份插手清单:

a = [####/boo ,#####/baa ,#####/bee ,####/bii , ####/buu]

编号#为4位数的随机编号。并且

b = [ boo , aaa , bii ]

我需要知道在名单上打字的条目是什么。页: 1 我得以完成这项工作,并同时使用in的操作器,对座标进行检查时,目前条目为B。但是,作为相对新鲜的y子,我几乎是正面的,这并不是书写它的最老或最容易的方法。因此,是否存在着减少我解决办法的这种区别?

Answer 1

The following code gives you an array with the indexes of a where the part after the slash is an element from b.

a_sep = [x.split( / )[1] for x in a]
idxs = [i for i, x in enumerate(a_sep) if x in b]

To improve performance, make b a set instead of a list.

解散:

>>> a = [ ####/boo ,  ####/baa ,  ####/bee ,  ####/bii ,  ####/buu ]
>>> b = [ boo ,  aaa ,  bii ]
>>> a_sep = [x.split( / )[1] for x in a]
>>> idxs = [i for i, x in enumerate(a_sep) if x in b]
>>> idxs
[0, 3]
>>> [a[i] for i in idxs]
[ ####/boo ,  ####/bii ]

如果你希望直接获得这些要素,而不是指数:

>>> a = [ ####/boo ,  ####/baa ,  ####/bee ,  ####/bii ,  ####/buu ]
>>> b = [ boo ,  aaa ,  bii ]
>>> [x for x in a if x.split( / )[1] in b]
[ ####/boo ,  ####/bii ]

Answer 2

ThiefMaster的回答是好的,地雷将非常相似,但如果你不了解指数,你可以做一个捷径:

>>> a = [ ####/boo ,  ####/baa ,  ####/bee ,  ####/bii ,  ####/buu ]
>>> b = [ boo ,  aaa ,  bii ]
>>> [x for x in a if x.split( / )[1] in b]
[ ####/boo ,  ####/bii ]

Again, if b is a set, that will improve performance for large numbers of elements.

Answer 3

import random
a=[str(random.randint(1000,9999))+ / +e for e in [ boo , baa , bee , bii , buu ]]

b = [ boo ,  aaa ,  bii ]

c=[x.split( / )[-1] for x in a if x.split( / )[-1] in b]

print c

prints:

[ boo ,  bii ]

或者,如果你想要全部入境:

print [x for x in a if x.split( / )[-1] in b]

prints:

[ 3768/boo ,  9110/bii ]

Answer 4

>>> [i for i in a for j in b if j in i]
[ ####/boo ,  ####/bii ]

我们应该做你们想要的、leg的和老的。

Answer 5

正如其他答复所示,你可以利用既定行动来更快地做到这一点。下面是这样做的一种方式:

>>> a_dict = dict((item.split( / )[1], item) for item in a)
>>> common = set(a_dict) & set(b)
>>> [a_dict[i] for i in common]
[ ####/boo ,  ####/bii ]

友情链接