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在列表上迭代格式字符串

language-comparisons iteration formatting python

Wayne Werner · 技术社区 · 14 年前

在Lisp中,您可以有如下内容:

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

在同一个列表上迭代的最蟒蛇式方法是什么?首先想到的是:

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])

但我觉得这很尴尬。我确定有更好的方法吗?

好吧,除非后来有人提供了一个更好的例子,我认为gnibler的解决方案是最好的\最接近的,尽管一开始可能不太明显它是如何做到的:

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

6 回复 | 直到 11 年前

John La Rooy 14 年前

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

或使用 .format

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

如果格式字符串不是硬编码的,您可以解析它来计算每行有多少个术语

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

综合起来就可以

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)

Josh Wright 14 年前

首先,我将使用2.6中更新的字符串格式化方法+

print "{0} {1} {2}".format(*mystuff[x:x+3])

Ken 14 年前

我想 join 是Python中最相似的功能:

(format t "~{~D, ~}" foo)

print(foo.join(", "))

当你有多个物品在里面的时候,情况会更糟,就像你看到的,但是如果你有一个 group-by 函数(无论如何,它确实有用!)我想你可以让它工作而不必太费事。类似:

mystuff = [1, 2, "Foo", 34, 42, "Ni", 12, 14, "Blue"]
print(["%d %d %s" % x for x in group(mystuff, 3)].join("\n"))

Walter Mundt 14 年前

我想说,最大的蟒蛇应该是把名单做得更深:

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

eyquem 11 年前

stuff = [1, 2, "Foo", 34, 42, "Ni", 12, 14, "Blue"]

it = iter(stuff)
itn = it.next

print '\n'.join("%d %d %s" % (el,itn(),itn())
                for el in it)

我觉得很容易理解

ksed 11 年前

基于赖特的两行:

mystuff = [1, 2, "Foo", 34, 42, "Ni", 12, 14, "Blue"]
print '\n'.join("{0},{1},{2}".format(*mystuff[x:x+3]) for x in xrange(0, len(mystuff)-1, 3))