python - 如何在python中通过正则表达式获取dict值

Question

dict1={'s1':[1,2,3],'s2':[4,5,6],'a':[7,8,9],'s3':[10,11]}

我怎样才能得到's'键的所有值？喜欢dict1['s*']得到的结果是dict1['s*']=[1,2,3,4,5,6,10,11]

Answer 1

>>> [x for d in dict1 for x in dict1[d] if d.startswith("s")]
[1, 2, 3, 4, 5, 6, 10, 11]

或者，如果它需要是一个正则表达式

>>> regex = re.compile("^s")
>>> [x for d in dict1 for x in dict1[d] if regex.search(d)]
[1, 2, 3, 4, 5, 6, 10, 11]

您在这里看到的是嵌套列表理解。相当于

result = []
for d in dict1:
    for x in dict1[d]:
        if regex.search(d):
            result.append(x)

因此，它的效率有点低，因为正则表达式的测试过于频繁（并且元素被一一附加）。所以另一个解决方案是

result = []
for d in dict1:
    if regex.search(d):
       result.extend(dict1[d])

Answer 2

>>> import re
>>> from itertools import chain
def natural_sort(l): 
     # http://stackoverflow.com/a/4836734/846892
     convert = lambda text: int(text) if text.isdigit() else text.lower() 
     alphanum_key = lambda key: [ convert(c) for c in re.split('([0-9]+)', key) ] 
     return sorted(l, key = alphanum_key)
... 

使用全局模式，'s*'：

>>> import fnmatch
def solve(patt):
    keys = natural_sort(k for k in dict1 if fnmatch.fnmatch(k, patt))
    return list(chain.from_iterable(dict1[k] for k in keys))
... 
>>> solve('s*')
[1, 2, 3, 4, 5, 6, 10, 11]

使用regex：

def solve(patt):
    keys = natural_sort(k for k in dict1 if re.search(patt, k))
    return list(chain.from_iterable( dict1[k] for k in keys ))
... 
>>> solve('^s')
[1, 2, 3, 4, 5, 6, 10, 11]

Answer 3

我在下面尝试了希望它对您有所帮助，方法是获取字典的键，然后如果键的第一个索引是您的字符开头，则使用键字典列表扩展列表。

n='s' # your start with character
result=[] # your output

for d in dict1.keys():
    if n == d[0]:
        result.extend(dict1[d])

print result