如何将打印输出格式化为固定宽度？

问题描述

$ p $ def $ splitter（str）：
（1，len（str））：
start = str [0：i]
end = str [i：]
yield（start，end）
for split splitter（end）：
result = [start]
result.extend（split）
yield result

el = [];

string =abcd
for splitter（abcd）：
el.extend（b）;

unique = sorted（set（el））;

作为唯一前缀：
如果前缀！=：
printvalue，前缀 - num = occurrences =，string.count（str ））;

我想打印所有在字符串变量中出现的排列。

由于排列的长度不一样，所以我想修正宽度并打印出来，不像这样：

 值a  - 出现次数= 1 
值ab  - 出现次数= 1 
值abc  - 出现次数= 1 
值b  -  num出现次数= 1 
值bc  - 出现次数= 1 
值bcd  - 出现次数= 1 
值c  - 出现次数= 1 
值cd  - 出现次数= 1 
值d  - 出现次数= 1 
  

如何使用格式来做到这一点？

我发现了这些帖子，但是对于字母数字字符串却不太合适：

解决方案

编辑12.11.2013 - 这个答案是非常古老的。它仍然是有效和正确的，但是看这个的人应该更喜欢。

您可以使用像这样：

 >>>打印'％5s'％'aa'
 aa 
>>>打印'％5s'％'aaa'
 aaa 
>>>打印'％5s'％'aaaa'
 aaaa 
>>>打印'％5s'％'aaaaa'
 aaaaa 
  

基本上：

• ％字符通知python，它必须将某些东西替换为一个标记。 $ b
• s 字符通知python令牌将是一个字符串
  
• 5
code>（或任何你想要的数字）通知python填充空格最多5个字符的字符串。




在你的一个可能的实现可能如下所示：



 >>> dict_ = {'a'：1，'ab'：1，'abc'：1} 
>>>对于dict_.items（）中的项目：
 ... print'value％3s  - 发生次数=％d'％item＃％d是整数标记
 ... 
值a发生次数= 1 
值发生次数= 1 
值abc  - 发生次数= 1 
  

边注：只是想知道您是否知道。例如，您可以在一行中获得所有组合的列表：



 >>> [''.join（perm）for i在范围内（1，len（s））为perm in it.permutations（s，i）] 
 ['a'，'b'，'c'，' d'，'ab'，'ac'，'ad'，'ba'，'bc'，'bd'，'ca'，'cb'，'cd'，'da'，'db'，'dc' ，abc，abd，acb，acd，adb，adc，bac，bad，bca，bcd，bda，bdc， cab，cad，cba，cbd，cda，cdb，dab，dac，dba，dbc，dca，dcb] b  

你可以通过使用组合结合 count（）。




I have this code (printing the occurrence of the all permutations in a string)

def splitter(str):
    for i in range(1, len(str)):
        start = str[0:i]
        end = str[i:]
        yield (start, end)
        for split in splitter(end):
            result = [start]
            result.extend(split)
            yield result

el =[];

string = "abcd"
for b in splitter("abcd"):
    el.extend(b);

unique =  sorted(set(el));

for prefix in unique:
    if prefix != "":
        print "value  " , prefix  , "- num of occurrences =   " , string.count(str(prefix));

I want to print all the permutation occurrence there is in string varaible.

since the permutation aren't in the same length i want to fix the width and print it in a nice not like this one:

value   a - num of occurrences =    1
value   ab - num of occurrences =    1
value   abc - num of occurrences =    1
value   b - num of occurrences =    1
value   bc - num of occurrences =    1
value   bcd - num of occurrences =    1
value   c - num of occurrences =    1
value   cd - num of occurrences =    1
value   d - num of occurrences =    1

How can I use format to do it?

I found these posts but it didn't go well with alphanumeric strings:

python string formatting fixed width

Setting fixed length with python

解决方案

EDIT 12.11.2013 - This answer is very old. It is still valid and correct, but people looking at this should prefer the new format syntax.

You can use string formatting like this:

>>> print '%5s' % 'aa'
   aa
>>> print '%5s' % 'aaa'
  aaa
>>> print '%5s' % 'aaaa'
 aaaa
>>> print '%5s' % 'aaaaa'
aaaaa

Basically:

• the % character informs python it will have to substitute something to a token
• the s character informs python the token will be a string
• the 5 (or whatever number you wish) informs python to pad the string with spaces up to 5 characters.

In your specific case a possible implementation could look like:

>>> dict_ = {'a': 1, 'ab': 1, 'abc': 1}
>>> for item in dict_.items():
...     print 'value %3s - num of occurances = %d' % item # %d is the token of integers
...
value   a - num of occurances = 1
value  ab - num of occurances = 1
value abc - num of occurances = 1

SIDE NOTE: Just wondered if you are aware of the existence of the itertools module. For example you could obtain a list of all your combinations in one line with:

>>> [''.join(perm) for i in range(1, len(s)) for perm in it.permutations(s, i)]
['a', 'b', 'c', 'd', 'ab', 'ac', 'ad', 'ba', 'bc', 'bd', 'ca', 'cb', 'cd', 'da', 'db', 'dc', 'abc', 'abd', 'acb', 'acd', 'adb', 'adc', 'bac', 'bad', 'bca', 'bcd', 'bda', 'bdc', 'cab', 'cad', 'cba', 'cbd', 'cda', 'cdb', 'dab', 'dac', 'dba', 'dbc', 'dca', 'dcb']

and you could get the number of occurrences by using combinations in conjunction with count().

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