在列表上迭代格式字符串

Question

在 Lisp 中，你可以有这样的东西：

(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？我首先想到的是：

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，除非后来有人提供更好的例子，否则我认为 gnibbler 的解决方案是最好\最接近的，尽管一开始可能不太明显它是如何做到这一点的：

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，您可以解析它以计算出每行有多少个术语

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

将它们放在一起给出

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

Answer 2

我认为 join 是 Python 中最相似的功能：

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

print(foo.join(", "))

正如你所见，当你里面有多个项目时，情况会更糟，尽管如果你有一个 group-by 函数（无论如何，这确实很有用！），我认为你可以使它的工作没有太多麻烦。像这样的东西：

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

Answer 3

对于初学者，我会使用 2.6+ 中较新的字符串格式化方法

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

Answer 4

我想说，最 Pythonic 的做法是让列表变得更深：

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)

我觉得非常好理解

Answer 6

基于赖特的两班轮：

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))