遍歷列表

Question

在Lisp中的格式字符串，可以有這樣的事情：

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

什麼是最Python的方式來遍歷同樣的名單？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]) 

但這只是感覺尷尬的我。我確定有更好的方法？

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那麼，除非後來有人提供了一個更好的例子，我覺得gnibbler的解決方案是最好的\最接近的，雖然它可能不會在第一它是如何做的事情確實是相當的明顯：

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) 

如果格式字符串不是硬編碼的，你可以分析它制定出每行

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

多少項把它放在一起給出

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

對於初學者來說，我會使用較新的字符串格式化方法在2.6+

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

Answer 3

我覺得join是Python中最相似的特徵：

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

print(foo.join(", ")) 

這是一個有點糟糕，當你有多個項目裏面，你看，但如果你有一個group-by功能（這是真的有用呢！），我想你可以把它沒有太多的麻煩工作。喜歡的東西：

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

我想說的最Python化將是使名單更深層次：基於賴特

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

Answer 5

兩個班輪：

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)) 

Answer 6

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) 

非常理解，我覺得