在列表中迭代格式字符串 [英] Iterate a format string over a list

问题描述

在Lisp中，你可以这样：

 （setf my-stuff'（1 2Foo34 42Ni12 14Blue））
（format t〜{〜d〜r〜s〜％〜}my-stuff）
  

在同一个列表上循环迭代的方式是什么？首先想到的是：

$ $ $ $ $ $ c $ mystuff = 1,2，Foo 34,42 Ni （0，len（mystuff）-1，3）：
print％d％d％s％tuple（mystuff [x， ：x + 3]）

但是这只是让我感到尴尬。我相信还有更好的办法吗？

---

好的，除非有人提供了一个更好的例子，我认为gnibbler的解决方案是最好的，最接近的，虽然它起初可能不是那么明显，它是如何做的：

  mystuff = [ 1，2，Foo，34，42，Ni，12,14，Blue] 
（* [iter（mystuff）] * 3）：
 print格式（* x）
  

解决方案 $ zip $（$ p code $ c $ mystuff = [1,2， * [iter（mystuff）] * 3）：
print％d $％$％$
$ c $ $ b

或使用 .format

  mystuff = [1 ，2，Foo，34,42，Ni，12,14，Blue] 
（* [iter（mystuff）] * 3）：
 print格式字符串（* x）
  

如果格式字符串不是硬编码，你可以解析它，以确定每行多少条款

 从条款g import formatter 
 num_terms = sum（1 for for x in Formatter（）。parse（{0} {1} {2}））
  

把它们放在一起给出

$ $ $ $ $ $ $ $ $ $ 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 zip（* [iter（mystuff）] * num_terms）：
print fmt.format（* x）

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)

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])

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

---

Well, unless someone later provides a better example, I think gnibbler's solution is the nicest\closest, 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

Or using .format

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

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}"))

Putting it all together gives

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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