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
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Answer
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.
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()
.