用零填充字母索引两侧的最简单方法是什么？答案

【问题标题】：Whats the easiest way to pad an alphabet index both sides with zeroes?用零填充字母索引两侧的最简单方法是什么？
【发布时间】：2014-04-30 15:08:03
【问题描述】：

我需要一个按以下格式对英语进行编码的数据集：
A = (1000000000000000000000000)
B = (0100000000000000000000000)
等等……

在 Python 中用零填充字母索引的最简单方法是什么？

到目前为止，我已经尝试过：

alphabet = ['A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z']

        for letter_index in range(len(alphabet)):
            print( '{0} = ({1})'.format(alphabet[letter_index], str(letter_index).rjust(letter_index).ljust(26-letter_index)))

但我的输出是这样的：

A = (0                         )
B = (1                        )
C = ( 2                      )
D = (  3                    )
E = (   4                  )
F = (    5                )
G = (     6              )
H = (      7            )
I = (       8          )
J = (        9        )
K = (        10      )
L = (         11    )
M = (          12  )
N = (           13)
O = (            14)
P = (             15)
Q = (              16)
R = (               17)
S = (                18)
T = (                 19)
U = (                  20)
V = (                   21)
W = (                    22)
X = (                     23)
Y = (                      24)
Z = (                       25)

最后，如果我想用逗号分隔数据集中的数字，最简单的方法是什么？
例如：
A = (1,0,0...)

【问题讨论】：

预期输出是什么？你的问题不是很清楚。
你是什么意思？预期的输出写在那里。我会这样标记它。
你的意思是像','.join(list('1000000000000000000000000'))这样的意思吗？你的问题我一点都不清楚。
@dawg -- 你甚至不需要在那里调用list -- 字符串也是可迭代的:)

标签： python padding

【解决方案1】：

以下代码将为您提供所需的输出：

alpha = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
zero25 = "0" * 25
for idx in range (len (alpha)):
    print "%s = (%s1%s)" % (alpha[idx], zero25[:idx], zero25[:25-idx])

具体来说，输出是：

A = (10000000000000000000000000)
B = (01000000000000000000000000)
C = (00100000000000000000000000)
D = (00010000000000000000000000)
E = (00001000000000000000000000)
F = (00000100000000000000000000)
G = (00000010000000000000000000)
H = (00000001000000000000000000)
I = (00000000100000000000000000)
J = (00000000010000000000000000)
K = (00000000001000000000000000)
L = (00000000000100000000000000)
M = (00000000000010000000000000)
N = (00000000000001000000000000)
O = (00000000000000100000000000)
P = (00000000000000010000000000)
Q = (00000000000000001000000000)
R = (00000000000000000100000000)
S = (00000000000000000010000000)
T = (00000000000000000001000000)
U = (00000000000000000000100000)
V = (00000000000000000000010000)
W = (00000000000000000000001000)
X = (00000000000000000000000100)
Y = (00000000000000000000000010)
Z = (00000000000000000000000001)

如果您希望它们用逗号分隔，则可以这样做：

alpha = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
zero25 = "0," * 25

print "%s = (1,%s)" % (alpha[0], zero25[:49])
for idx in range (1, len (alpha)):
    print "%s = (%s,1%s)" % (alpha[idx], zero25[:idx*2-1], zero25[idx*2-1:-1])

输出：

A = (1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
B = (0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
C = (0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
D = (0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
E = (0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
F = (0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
G = (0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
H = (0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
I = (0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
J = (0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
K = (0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
L = (0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
M = (0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0)
N = (0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0)
O = (0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0)
P = (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0)
Q = (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0)
R = (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0)
S = (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0)
T = (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0)
U = (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0)
V = (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0)
W = (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0)
X = (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0)
Y = (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0)
Z = (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1)

【讨论】：

【解决方案2】：

你可以做类似的事情：

>>> '{:0<26}'.format('1'.zfill(1))
'10000000000000000000000000'
>>> '{:0<26}'.format('1'.zfill(2))
'01000000000000000000000000'

一个完整的例子是这样的：

>>> c = 'A'
>>> '{} = ({:0<26})'.format(c, '1'.zfill(ord(c)-ord('A')+1))
'A = (10000000000000000000000000)'
>>> c = 'B'
>>> '{} = ({:0<26})'.format(c, '1'.zfill(ord(c)-ord('A')+1))
'B = (01000000000000000000000000)'
>>> c = 'C'
>>> '{} = ({:0<26})'.format(c, '1'.zfill(ord(c)-ord('A')+1))
'C = (00100000000000000000000000)'

【讨论】：

【解决方案3】：

>>> for i in range(26):
...   print "{} = ({}1{})".format(chr(i+65),'0'*i,'0'*(25-i))
... 
A = (10000000000000000000000000)
B = (01000000000000000000000000)
C = (00100000000000000000000000)
D = (00010000000000000000000000)
E = (00001000000000000000000000)
F = (00000100000000000000000000)
G = (00000010000000000000000000)
H = (00000001000000000000000000)
I = (00000000100000000000000000)
J = (00000000010000000000000000)
K = (00000000001000000000000000)
L = (00000000000100000000000000)
M = (00000000000010000000000000)
N = (00000000000001000000000000)
O = (00000000000000100000000000)
P = (00000000000000010000000000)
Q = (00000000000000001000000000)
R = (00000000000000000100000000)
S = (00000000000000000010000000)
T = (00000000000000000001000000)
U = (00000000000000000000100000)
V = (00000000000000000000010000)
W = (00000000000000000000001000)
X = (00000000000000000000000100)
Y = (00000000000000000000000010)
Z = (00000000000000000000000001)

【讨论】：