Python：将函数列表应用于列表中的每个元素

Say I have list with elements content = ['121\n', '12\n', '2\n', '322\n'] and list with functions fnl = [str.strip, int].

假设我有元素content = ['121 \ n','12 \ n','2 \ n','322 \ n']列表,并列出函数fnl = [str.strip,int]。

So I need to apply each function from fnl to each element from content sequentially. I can do this by several calls map.

所以我需要将fnl中的每个函数依次应用于内容中的每个元素。我可以通过几个调用map来做到这一点。

Another way:

xl = lambda func, content: map(func, content)
for func in fnl:
    content = xl(func, content)

I'm just wondering if there is a more pythonic way to do it.

我只是想知道是否有更多的pythonic方式来做到这一点。

Without separate function? By single expression?

没有单独的功能?通过单一表达?

2 个解决方案

#1

You could use the reduce() function in a list comprehension here:

您可以在列表推导中使用reduce()函数:

[reduce(lambda v, f: f(v), fnl, element) for element in content]

Demo:

>>> content = ['121\n', '12\n', '2\n', '322\n']
>>> fnl = [str.strip, int]
>>> [reduce(lambda v, f: f(v), fnl, element) for element in content]
[121, 12, 2, 322]

This applies each function in turn to each element, as if you nested the calls; for fnl = [str.strip, int] that translates to int(str.strip(element)).

这会依次将每个函数应用于每个元素,就像嵌套调用一样; for fnl = [str.strip,int]转换为int(str.strip(element))。

In Python 3, reduce() was moved to the functools module; for forwards compatibility, you can import it from that module from Python 2.6 onwards:

在Python 3中,reduce()被移动到functools模块;为了向前兼容,您可以从Python 2.6开始从该模块导入它:

from functools import reduce

results = [reduce(lambda v, f: f(v), fnl, element) for element in content]

Note that for the int() function, it doesn't matter if there is extra whitespace around the digits; int('121\n') works without stripping of the newline.

请注意,对于int()函数,数字周围是否有额外的空格并不重要; int('121 \ n')可以在不删除换行符的情况下工作。

#2

You are describing the basic use of a list comprehension:

您正在描述列表理解的基本用法:

>>> content = ['121\n', '12\n', '2\n', '322\n']
>>> [int(n) for n in content]
[121, 12, 2, 322]

Note you don't need the call to strip to convert to integer here, some whitespace is handled fine.

请注意,这里不需要调用strip来转换为整数,一些空格处理得很好。

If your real use-case is more complex and you wish to compose arbitrarily many functions in the comprehension, however, I found the idea from here quite pythonic:

如果您的真实用例更复杂,并且您希望在理解中任意编写许多函数,那么,我发现这里的想法非常pythonic:

def compose(f1, f2):
    def composition(*args, **kwargs):
        return f1(f2(*args, **kwargs))
    return composition

def compose_many(*funcs):
    return reduce(compose, funcs)

#1