循环打印两个列表,以获得两列在每个列表的每个元素的第一个字母之间具有固定(自定义集)空间

Question

假设我有这两个列表:

column1 = ["soft","pregnant","tall"]
column2 = ["skin","woman", "man"]

如何使用自定义的固定空间(例如10,如示例)循环打印这两个列表,从第一个列表的每个元素的第一个字母开始,直到第二个列表的每个元素的第一个字母？

设置间距为10的示例输出:

soft      skin
pregnant  woman 
tall      man

Answer 1

轻松完成字符串格式化,

column1 = ["soft","pregnant","tall"]
column2 = ["skin","woman", "man"]

for c1, c2 in zip(column1, column2):
    print "%-9s %s" % (c1, c2)

或者您可以使用str.ljust,如果您希望填充基于变量,则更整洁:

padding = 9
for c1, c2 in zip(column1, column2):
    print "%s %s" % (c1.ljust(padding), c2)

(注意:填充9不是10因为单词之间的硬编码空间)

Answer 2

怎么样:

>>> column1 = ["soft","pregnant","tall"]
>>> column2 = ["skin","woman", "man"]
>>> for line in zip(column1, column2):
...     print '{:10}{}'.format(*line)
... 
soft      skin
pregnant  woman
tall      man