Python: how to replace an element in a list with all elements from another list?

Question

I have two lists in Python that look like this:

 lst = [1, '?2']
 replace_lst1 = ['a','b','c']

For each occurrence of '?2' in lst, I would like to replace it with each element from replace_lst1 thereby producing a list of lists as follows:

res = [ [1,'a'], 
        [1,'b'],
        [1,'c'] ]

Similarly, if I have the following lists:

lst = [1, '?2','?3']
replace_lst1 = ['a','b','c']
replace_lst2 = ['A', 'B', 'C']

I would like to replace '?2' with each element from replace_lst1 and '?3' with each element from replace_lst2, thereby exploring all possible permutations. The result should look like this:

res = [ [1,'a','A'], 
        [1,'a','B'],
        [1,'a','C'],
        [1,'b','A'], 
        [1,'b','B'],
        [1,'b','C'],
        [1,'c','A'], 
        [1,'c','B'],
        [1,'c','C'] ]

It would be great if you could provide me with some suggestions how to proceed.

Thanks!

what is the purpose of this? Do you ever have replace_lst3? — jamylak
– jamylak, Commented Jun 6, 2013 at 11:58
are the ?N format mandatory? shall it begin at 1? it's definitively possible to do what you want, but it looks pretty weird and won't be elegant... — zmo
– zmo, Commented Jun 6, 2013 at 12:01
Unless this is being imposed on you from some external place, I would strongly advise you to use a different data model. What you want can be done very elegantly with a single line of code, but not if you define your data like this. — Tim Pietzcker
– Tim Pietzcker, Commented Jun 6, 2013 at 12:01
In this context, permutations has a specific meaning, and it isn't what you want — John La Rooy
– John La Rooy, Commented Jun 6, 2013 at 12:17
There is no replace_lst3. The ?X format is given. It doesn't begin at 1, actually, the lists are strings. I simplified the things for illustrative purposes. As for the data structure, I'd like the result to be a list of lists since I'll have to do some operations on the result and with this data structure, I know how to proceed. — katyaa
– katyaa, Commented Jun 6, 2013 at 12:19

Tim Pietzcker · Accepted Answer · 2013-06-06 12:07:10Z

3

If you change your data structure slightly, this is a trivial problem for the itertools module:

>>> lst = [1]
>>> combine = [["a", "b", "c"], ["A", "B", "C"]]
>>> import itertools
>>> [lst+list(item) for item in itertools.product(*combine)]
[[1, 'a', 'A'], [1, 'a', 'B'], [1, 'a', 'C'], [1, 'b', 'A'], [1, 'b', 'B'], 
 [1, 'b', 'C'], [1, 'c', 'A'], [1, 'c', 'B'], [1, 'c', 'C']]

answered Jun 6, 2013 at 12:07

Tim Pietzcker

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John La Rooy · Accepted Answer · 2013-06-06 12:16:31Z

3

>>> from itertools import product
>>> lst = [1]
>>> combine = [["a", "b", "c"], ["A", "B", "C"]]
>>> list(product(*[lst]+combine))
[(1, 'a', 'A'), (1, 'a', 'B'), (1, 'a', 'C'), (1, 'b', 'A'), (1, 'b', 'B'), (1, 'b', 'C'), (1, 'c', 'A'), (1, 'c', 'B'), (1, 'c', 'C')]

you could also use

list(product(lst, replace_lst1, replace_lst2))

edited Jun 6, 2013 at 12:16

answered Jun 6, 2013 at 12:10

John La Rooy

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njzk2 · Accepted Answer · 2013-06-06 12:22:38Z

0

I would use itertools.product, plus a test to replace the ? by the value :

lst = [1, '?2','?3']
replace_lst1 = ['a','b','c']
replace_lst2 = ['A', 'B', 'C']
res = []
#put as many replace_lst as you need here
for values in itertools.product(replace_lst1, replace_lst2):
    val_iter = iter(values)
    res.append([x if str(x).find('?') == -1 else next(val_iter) for x in lst])

The use of val_iter allows for the ?* to be placed anywhere (but not in any order, though).

answered Jun 6, 2013 at 12:22

njzk2

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1 Comment

katyaa Over a year ago

Thank you all for the responses! itertools.product is exactly what I needed! Cheers!

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