nested looping in python

Question

in a string suppose 12345 , i want to take nested loops , so that i would be able to iterate through the string in this following way :-

1, 2, 3, 4, 5 would be taken as integers
12, 3, 4,5 as integers
1, 23, 4, 5 as integers
1, 2, 34, 5 as integers ...

And so on. I know what's the logic but being a noob in Python, I'm not able to form the loop.

So if you know the logic, perhaps you should try describing it in words, as a complement to the example output. — unwind
– unwind, Commented Dec 15, 2009 at 13:42
So bascially you're looking for an algorithm that yields 1. the 1 possibility to omit no commas in your array 2. the 4 possibilities to omit one comma 3. the 6 possibilities to omit two commas 4. the 4 possibilities to omit three commas 5. the 1 possibility to omit four commas ? — Johannes Charra
– Johannes Charra, Commented Dec 15, 2009 at 14:10

Mike D. · Accepted Answer · 2009-12-15 13:43:00Z

2

This smells a bit like homework.

Try writing down the successive outputs, one per line, and look for a pattern. See if you can explain that pattern with slices of the input string. Then look for a numeric pattern to the slicing.

Also, please edit your question to put quotes around your strings. What you've written isn't very clear in terms of the outputs, whether you output strings with commas or lists of substrings.

answered Dec 15, 2009 at 13:43

Mike D.

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Michal Čihař · Accepted Answer · 2009-12-15 13:46:34Z

1

You can do the inner traversals by following code, the first traversal is trivial.

s = '12345'

chars = [c for c in s]

for i in range(len(s) - 1):
    print '%d:' % i,
    for el in chars[:i] + [chars[i] + chars[i + 1]] + chars[i + 2:]:
        print el,
    print

answered Dec 15, 2009 at 13:46

Michal Čihař

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Comments

jsbueno · Accepted Answer · 2009-12-15 13:46:52Z

1

number = 12345

str_number = str(number)

output = []
for index, part in enumerate(str_number[:-1]):
    output_part = []
    for second_index, second_part in enumerate(str_number):
        if index == second_index:
            continue
        elif index == second_index - 1:
            output_part.append(int(part + second_part))
        else:
            output_part.append(int(second_part))
    output.append(output_part)

print output

STick it inside a function definition and put an "yield output_part" in place of the "output.append" line to get a usefull interator.

answered Dec 15, 2009 at 13:46

jsbueno

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