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问题描述

我有这段代码(将所有排列的出现打印在字符串中)

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));

我要打印字符串可变项中所有出现的排列.

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

因为排列的长度不同,所以我想固定宽度并以一种不错的方式打印它,而不是这样:

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

如何使用format来做到这一点?

How can I use format to do it?

我找到了这些帖子,但与字母数字字符串的配合不太好:

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

python字符串格式固定宽度

使用python设置固定长度

推荐答案

编辑2013-12-11 -此答案非常老.它仍然是正确和正确的,但是关注此问题的人们应该更喜欢新格式语法.

EDIT 2013-12-11 - This answer is very old. It is still valid and correct, but people looking at this should prefer the new format syntax.

您可以像这样使用字符串格式:

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

基本上:

  • %字符通知python,它将必须用某些东西代替令牌
  • s字符通知python令牌将是一个字符串
  • 5(或您想要的任何数字)会通知python在字符串中填充最多5个字符.
  • 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.

在您的特定情况下,可能的实现方式如下:

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

侧注::想知道您是否知道 itertools模块.例如,您可以使用以下命令在一行中获取所有组合的列表:

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']

,通过将combinationscount()结合使用,您可以得到出现的次数.

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

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09-03 09:46