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， str.ljust整洁：

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

（注意：填充是9而不是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

Answer 3

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

for row in zip(column1, column2):
    print "%-9s %s" % row # formatted to a width of 9 with one extra space after

Answer 4

使用Python 3

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

for line in zip(column1, column2):
    print('{:10}{}'.format(*line))

Answer 5

使用新样式字符串格式的一个班轮：

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

>>> print "\n".join("{0}\t{1}".format(a, b) for a, b in zip(column1, column2))

soft        skin
pregnant    woman
tall        man