Replacing list item with contents of another list

Question

Similar to this question, but instead of replacing one item with another, I\'d like to replace any occurrences of one item with the contents of a list.

orig          


        

Answer 1

No need for anything fancy:

desired = orig[:2] + repl + orig[3:]

To find 2 you can search for orig.index('c').

x = orig.index('c')
desired = orig[:x] + repl + orig[x+1:]

if repl is not a list, just use list(repl)

Answer 2

Different approach: when I'm doing replacements, I prefer to think in terms of dictionaries. So I'd do something like

>>> orig = [ 'a', 'b', 'c', 'd' ]
>>> rep = {'c': ['x', 'y', 'z']}
>>> [i for c in orig for i in rep.get(c, [c])]
['a', 'b', 'x', 'y', 'z', 'd']

where the last line is the standard flattening idiom.

One advantage (disadvantage?) of this approach is that it'll handle multiple occurrences of 'c'.

[update:]

Or, if you prefer:

>>> from itertools import chain
>>> list(chain.from_iterable(rep.get(c, [c]) for c in orig))
['a', 'b', 'x', 'y', 'z', 'd']

On the revised test case:

>>> orig = [ 'a', 'b', 'c', 'd', 'c' ]
>>> rep = {'c': ['x', 'y', 'z']}
>>> list(chain.from_iterable(rep.get(c, [c]) for c in orig))
['a', 'b', 'x', 'y', 'z', 'd', 'x', 'y', 'z']

Answer 3

>>> orig = [ 'a', 'b', 'c', 'd' ]
>>> repl = [ 'x', 'y', 'z' ]
>>> desired = list(orig)  #can skip this and just use `orig` if you don't mind modifying it (and it is a list already)
>>> desired[2:3] = repl
>>> desired
['a', 'b', 'x', 'y', 'z', 'd']

And of course, if you don't know that 'c' is at index 2, you can use orig.index('c') to find out that information.

Answer 4

If you enumerate backwards, you can extend the list as you go because the items you move have already gone through the enumeration.

>>> orig = [ 'a', 'b', 'c', 'd', 'c' ]
>>> repl = [ 'x', 'y', 'z' ]
>>> desired = [ 'a', 'b', 'x', 'y', 'z', 'd', 'x', 'y', 'z' ]
>>> for i in xrange(len(orig)-1, -1, -1):
...     if orig[i] == 'c':
...             orig[i:i+1] = repl
... 
>>> orig
['a', 'b', 'x', 'y', 'z', 'd', 'x', 'y', 'z']

Answer 5

Yet another way:

>>> import operator
>>> orig = [ 'a', 'b', 'c', 'd', 'c' ]
>>> repl = [ 'x', 'y', 'z' ]
>>> output = [repl if x == 'c' else [x] for x in orig]
>>> reduce(operator.add, output)
['a', 'b', 'x', 'y', 'z', 'd', 'x', 'y', 'z']
>>>