在列表上迭代格式字符串

2022-01-24 00:00:00 python iteration formatting language-comparisons

问题描述

在 Lisp 中，你可以有这样的东西:

In Lisp, you can have something like this:

(setf my-stuff '(1 2 "Foo" 34 42 "Ni" 12 14 "Blue"))
(format t "~{~d ~r ~s~%~}" my-stuff)

迭代同一个列表的最 Pythonic 方式是什么?首先想到的是:

What would be the most Pythonic way to iterate over that same list? The first thing that comes to mind is:

mystuff = [1, 2, "Foo", 34, 42, "Ni", 12, 14, "Blue"]
for x in xrange(0, len(mystuff)-1, 3):
    print "%d %d %s" % tuple(mystuff[x:x+3])

但这对我来说感觉很尴尬.我确定有更好的方法吗?

But that just feels awkward to me. I'm sure there's a better way?

好吧，除非有人后来提供了更好的例子，否则我认为 gnibbler 的解决方案是最好的最接近的，尽管起初它的工作方式可能并不那么明显:

Well, unless someone later provides a better example, I think gnibbler's solution is the nicestclosest, though it may not be quite as apparent at first how it does what it does:

mystuff = [1, 2, "Foo", 34, 42, "Ni", 12, 14, "Blue"]
for x in zip(*[iter(mystuff)]*3):
    print "{0} {1} {2}".format(*x)

解决方案

mystuff = [1, 2, "Foo", 34, 42, "Ni", 12, 14, "Blue"]
for x in zip(*[iter(mystuff)]*3):
    print "%d %d %s"%x

或者使用.format

mystuff = [1, 2, "Foo", 34, 42, "Ni", 12, 14, "Blue"]
for x in zip(*[iter(mystuff)]*3):
    print "{0} {1} {2}".format(*x)

如果格式字符串没有硬编码，您可以对其进行解析以计算出每行有多少项

If the format string is not hardcoded, you can parse it to work out how many terms per line

from string import Formatter
num_terms = sum(1 for x in Formatter().parse("{0} {1} {2}"))

把它们放在一起给出了

mystuff = [1, 2, "Foo", 34, 42, "Ni", 12, 14, "Blue"]
fmt = "{0} {1} {2}"
num_terms = sum(1 for x in Formatter().parse(fmt))
for x in zip(*[iter(mystuff)]*num_terms):
    print fmt.format(*x)

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