Python检查列表项中的字符串

Question

例如，问题如下：

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

因此，我需要从列表“ lst”中获取字典“ d”的键中包含的所有元素，而没有子字符串匹配，因为如果执行“ print d [i]”，则会出现错误。

Answer 1

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

Answer 2

这行应该做的工作：

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

与您的数据：

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

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

Answer 3

使用sets ：

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

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

Answer 4

>>> 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']