python - 如何将可变数量的参数格式化为字符串？
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python - How to format variable number of arguments into a string?
提问by Gerard
We know that formatting oneargument can be done using one%sin a string:
我们知道，格式化一个参数是可以做到用一个%s在一个字符串：
>>> "Hello %s" % "world"
'Hello world'
for two arguments, we can use two %s(duh!):
对于两个参数，我们可以使用两个%s（废话！）：
>>> "Hello %s, %s" % ("John", "Joe")
'Hello John, Joe'
So, how can I format a variable number of arguments without having to explicitly define within the base string a number of %sequal to the number of arguments to format? it would be very cool if something like this exists:
那么，如何格式化可变数量的参数，而不必在基本字符串中显式定义%s与 format 的参数数量相等的数量？如果存在这样的东西，那就太酷了：
>>> "Hello <cool_operator_here>" % ("John", "Joe", "Mary")
Hello JohnJoeMary
>>> "Hello <cool_operator_here>" % ("John", "Joe", "Mary", "Rick", "Sophie")
Hello JohnJoeMaryRickSophie
Is this even possible or the only thing I could do about it is to do something like:
这是否可能，或者我唯一能做的就是做这样的事情：
>>> my_args = ["John", "Joe", "Mary"]
>>> my_str = "Hello " + ("".join(["%s"] * len(my_args)))
>>> my_str % tuple(my_args)
"Hello JohnJoeMary"
NOTE: I needto do it with the %sstring formatting operator.
注意：我需要使用%s字符串格式化操作符来完成。
UPDATE:
更新：
It needs to be with the %sbecause a function from another library formats my string using that operator given that I pass the unformatted string and the args to format it, but it makes some checking and corrections (if needed) on the args before actually making the formatting.
它需要与 ，%s因为来自另一个库的函数使用该运算符格式化我的字符串，因为我传递了未格式化的字符串和 args 来格式化它，但它在实际制作之前对 args 进行了一些检查和更正（如果需要）格式化。
So I need to call it:
所以我需要调用它：
>>> function_in_library("Hello <cool_operator_here>", ["John", "Joe", "Mary"])
"Hello JohnJoeMary"
Thanks for your help!
谢谢你的帮助！
采纳答案by Martijn Pieters
You'd use str.join()on the list withoutstring formatting, then interpolate the result:
您可以str.join()在没有字符串格式的列表上使用，然后插入结果：
"Hello %s" % ', '.join(my_args)
Demo:
演示：
>>> my_args = ["foo", "bar", "baz"]
>>> "Hello %s" % ', '.join(my_args)
'Hello foo, bar, baz'
If some of your arguments are not yet strings, use a list comprehension:
如果您的某些参数还不是字符串，请使用列表推导式：
>>> my_args = ["foo", "bar", 42]
>>> "Hello %s" % ', '.join([str(e) for e in my_args])
'Hello foo, bar, 42'
or use map(str, ...):
或使用map(str, ...)：
>>> "Hello %s" % ', '.join(map(str, my_args))
'Hello foo, bar, 42'
You'd do the same with your function:
你会对你的函数做同样的事情：
function_in_library("Hello %s", ', '.join(my_args))
If you are limited by a (rather arbitrary) restriction that you cannot use a joinin the interpolation argument list, use a jointo create the formatting string instead:
如果您受到不能join在插值参数列表中使用 a的（相当任意的）限制，请改用 ajoin创建格式化字符串：
function_in_library("Hello %s" % ', '.join(['%s'] * len(my_args)), my_args)
