Question

在Lisp中，你可以这样：

(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])

但这对我来说感觉很尴尬。我确定有更好的方法吗？

好吧，除非有人后来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，虽然它起初可能不那么明显它的作用：

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

Answer 1

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)

如果格式字符串没有硬编码，您可以解析它以计算每行的条件数

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

将所有内容放在一起会产生

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)

Answer 2

对于初学者，我会在2.6 +

中使用较新的字符串格式化方法

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

Answer 3

我认为join是Python中最相似的功能：

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

print(foo.join(", "))

如你所见，当你有多个项目时，情况会更糟，但如果你有一个group-by功能（这真的很有用！），我认为你可以毫无困难地使它工作。类似的东西：

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

Answer 4

我认为Pythonic最多的是让名单更深入：

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

Answer 5

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)

非常可以理解，我认为

Answer 6

基于赖特的双人班轮：

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))

在列表上迭代格式字符串

6 个答案: