Python检查字符串列表项

时间:2013-04-23 08:51:58

标签: python

问题如下:例如:

lst = ['2456', '1871', '187']

d = {
    '1871': '1',
    '2456': '0',
}

for i in lst:
    if any(i in x for x in d.keys()):
        print i

% python test.py
2456
1871
187

所以,我需要从字典“d”的键中包含列表“lst”中的所有元素,而没有子字符串匹配,因为如果我执行“print d [i]”,我得到一个错误。

4 个答案:

答案 0 :(得分:3)

>>> lst = ['2456', '1871', '187']
>>> d = {
    '1871': '1',
    '2456': '0',
}
>>> [x for x in lst if x in d]
['2456', '1871']

答案 1 :(得分:1)

这一行应该完成这项工作:

 l=[e for e in d if e in lst]

包含您的数据:

In [5]: l=[e for e in d if e in lst]

In [6]: l
Out[7]: ['2456', '1871']

答案 2 :(得分:0)

使用sets

lst = ['2456', '1871', '187']
d = {'1871': '1', '2456': '0'}

print(set(lst) & set(d.keys())) # prints '{'2456', '1871'}'

答案 3 :(得分:0)

>>> for i in li:
    if i in d:
        print "{0} => {1}".format(i,d[i])


2456 => 0
1871 => 1

在列表理解中:

>>> [i for i in li if i in d]
['2456', '1871']