Python 检查字符串是否以列表中的字符串之一结尾
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Check if string ends with one of the strings from a list
提问by TheMeaningfulEngineer
What is the pythonic way of writing the following code?
编写以下代码的pythonic方式是什么？
extensions = ['.mp3','.avi']
file_name = 'test.mp3'
for extension in extensions:
if file_name.endswith(extension):
#do stuff
I have a vague memory that the explicit declaration of the forloop can be avoided and be written in the ifcondition. Is this true?
我有一个模糊的记忆，for可以避免循环的显式声明并将其写入if条件中。这是真的？
采纳答案by falsetru
Though not widely known, str.endswithalso accepts a tuple. You don't need to loop.
虽然并不广为人知，但str.endswith也接受一个元组。你不需要循环。
>>> 'test.mp3'.endswith(('.mp3', '.avi'))
True
回答by Jon Clements
Just use:
只需使用：
if file_name.endswith(tuple(extensions)):
回答by alecxe
Take an extension from the file and see if it is in the set of extensions:
从文件中取出一个扩展名，看看它是否在扩展名集中：
>>> import os
>>> extensions = set(['.mp3','.avi'])
>>> file_name = 'test.mp3'
>>> extension = os.path.splitext(file_name)[1]
>>> extension in extensions
True
Using a set because time complexity for lookups in sets is O(1) (docs).
使用集合是因为在集合中查找的时间复杂度是 O(1) ( docs)。
回答by Thomas Wouters
I have this:
我有这个：
def has_extension(filename, extension):
ext = "." + extension
if filename.endswith(ext):
return True
else:
return False
回答by Igor A
There is two ways: regular expressions and string (str) methods.
有两种方法：正则表达式和字符串（str）方法。
String methods are usually faster ( ~2x ).
字符串方法通常更快（ ~2x ）。
import re, timeit
p = re.compile('.*(.mp3|.avi)$', re.IGNORECASE)
file_name = 'test.mp3'
print(bool(t.match(file_name))
%timeit bool(t.match(file_name)
792 ns ± 1.83 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
每个循环 792 ns ± 1.83 ns（7 次运行的平均值 ± 标准偏差，每次 1000000 次循环）
file_name = 'test.mp3'
extensions = ('.mp3','.avi')
print(file_name.lower().endswith(extensions))
%timeit file_name.lower().endswith(extensions)
274 ns ± 4.22 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
每个循环 274 ns ± 4.22 ns（7 次运行的平均值 ± 标准偏差，每次 1000000 次循环）
回答by Xxxo
I just came across this, while looking for something else.
我只是在寻找其他东西时遇到了这个。
I would recommend to go with the methods in the ospackage. This is because you can make it more general, compensating for any weird case.
我建议使用os包中的方法。这是因为您可以使它更通用，以补偿任何奇怪的情况。
You can do something like:
您可以执行以下操作：
import os
the_file = 'aaaa/bbbb/ccc.ddd'
extensions_list = ['ddd', 'eee', 'fff']
if os.path.splitext(the_file)[-1] in extensions_list:
# Do your thing.
回答by NeverHopeless
Another possibility could be to make use of IN statement:
另一种可能性是使用 IN 语句：
extensions = ['.mp3','.avi']
file_name = 'test.mp3'
if "." in file_name and file_name[file_name.rindex("."):] in extensions:
print(True)
回答by Akash Singh
another way which can return the list of matching strings is
可以返回匹配字符串列表的另一种方法是
sample = "alexis has the control"
matched_strings = filter(sample.endswith, ["trol", "ol", "troll"])
print matched_strings
['trol', 'ol']
