Replacing a list item with the contents of another list

Question

Like this question , but instead of replacing one element with another, I would like to replace any occurrences of one element with the contents of the list.

orig = [ 'a', 'b', 'c', 'd', 'c' ]
repl = [ 'x', 'y', 'z' ]
desired = [ 'a', 'b', 'x', 'y', 'z', 'd', 'x', 'y', 'z' ]

# these are all incorrect, or fail to compile
[ repl if x == 'c' else x for x in orig ]
[ [a for a in orig] if x == 'c' else x for x in orig ]
[ (a for a in orig) if x == 'c' else x for x in orig ]
[ a for a in orig if x == 'c' else x for x in orig ]

Edit: I realized that I wanted to replace all occurrences of an element, not just the first one. (Apologizes to those who did not cover this matter in their reply.)

Answer 1

A different approach: when I make replacements, I prefer to think in terms of dictionaries. So I would 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 anti-aliasing idiom.

(?) , 'c'.

[:]

, :

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

:

>>> 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 2

>>> 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']

, , , 'c' 2, orig.index('c'), .

Answer 3

:

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

2, orig.index('c').

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

repl , list(repl)

Answer 4

, , , , , .

>>> 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

:

>>> 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']
>>>