我有一个列表列表,其中包含代表卡片中卡片的字符串。我需要垂直打印列表。
stacks = [['4C', 'QD', '9D', 'TD', 'JH', 'JD', '6C', 'AC', '3C', 'TS', 'TH'], ['AH', '2C', 'KD', '8C', '5C', '5H', '2D', '5D', '7C', 'AS', 'AD'], ['3S', 'KH', '3H', 'TC', 'QS', '4D', '7S', '7H', '9C', '4S'], ['JS', '7D', 'KC', 'QH', '6S', '6H', '8H', '8S', 'JC', '5S'], ['KS', '2S', 'QC', '4H', '6D', '9S', '2H', '8D', '3D', '9H']]
我需要它打印如下:
4C AH 3S JS KS
QD 2C KH 7D 2S
继续这样
答案 0 :(得分:1)
您可以使用zip
:
stacks = [['4C', 'QD', '9D', 'TD', 'JH', 'JD', '6C', 'AC', '3C', 'TS', 'TH'], ['AH', '2C', 'KD', '8C', '5C', '5H', '2D', '5D', '7C', 'AS', 'AD'], ['3S', 'KH', '3H', 'TC', 'QS', '4D', '7S', '7H', '9C', '4S'], ['JS', '7D', 'KC', 'QH', '6S', '6H', '8H', '8S', 'JC', '5S'], ['KS', '2S', 'QC', '4H', '6D', '9S', '2H', '8D', '3D', '9H']]
stacks = zip(*stacks)
print stacks
<强>输出强>
[('4C', 'AH', '3S', 'JS', 'KS'),
('QD', '2C', 'KH', '7D', '2S'),
('9D', 'KD', '3H', 'KC', 'QC'),
('TD', '8C', 'TC', 'QH', '4H'),
('JH', '5C', 'QS', '6S', '6D'),
('JD', '5H', '4D', '6H', '9S'),
('6C', '2D', '7S', '8H', '2H'),
('AC', '5D', '7H', '8S', '8D'),
('3C', '7C', '9C', 'JC', '3D'),
('TS', 'AS', '4S', '5S', '9H')]
答案 1 :(得分:1)
前两行比其他行长,因此请使用itertools.izip_longest()
代替标准zip()
来保留该数据。我使用了填充值''
,用于不等长度列表中的缺失值。
from itertools import izip_longest
stacks = [['4C', 'QD', '9D', 'TD', 'JH', 'JD', '6C', 'AC', '3C', 'TS', 'TH'], ['AH', '2C', 'KD', '8C', '5C', '5H', '2D', '5D', '7C', 'AS', 'AD'], ['3S', 'KH', '3H', 'TC', 'QS', '4D', '7S', '7H', '9C', '4S'], ['JS', '7D', 'KC', 'QH', '6S', '6H', '8H', '8S', 'JC', '5S'], ['KS', '2S', 'QC', '4H', '6D', '9S', '2H', '8D', '3D', '9H']]
for row in izip_longest(*stacks, fillvalue=''):
print ' '.join(row)
输出:
4C AH 3S JS KS
QD 2C KH 7D 2S
9D KD 3H KC QC
TD 8C TC QH 4H
JH 5C QS 6S 6D
JD 5H 4D 6H 9S
6C 2D 7S 8H 2H
AC 5D 7H 8S 8D
3C 7C 9C JC 3D
TS AS 4S 5S 9H
TH AD
如果你不想要最后一行的迷路空间:
for row in izip_longest(*stacks):
print ' '.join(col for col in row if col is not None)
这是一种只使用嵌套循环的方法:
longest = max(len(row) for row in stacks)
for i in range(longest):
for row in stacks:
if i < len(row):
print row[i],
print
答案 2 :(得分:0)
使用zip转置列表:
for vlist in zip(*stacks):
print ' '.join(vlist)