how to get dict value by regex in python
dictionary, python, regex
Solution
>>> [x for d in dict1 for x in dict1[d] if d.startswith("s")]
[1, 2, 3, 4, 5, 6, 10, 11]
or, if it needs to be a regex
>>> 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]
What you're seeing here is a nested list comprehension. It's equivalent to
result = []
for d in dict1:
for x in dict1[d]:
if regex.search(d):
result.append(x)
As such, it's a little inefficient because the regex is tested way too often (and the elements are appended one by one). So another solution would be
result = []
for d in dict1:
if regex.search(d):
result.extend(dict1[d])
Problem
``` dict1={'s1':[1,2,3],'s2':[4,5,6],'a':[7,8,9],'s3':[10,11]} ``` how can I get all the value which key is with 's'? like `dict1['s*']`to get the result is `dict1['s*']=[1,2,3,4,5,6,10,11]`