How to replace a string inside a list with a sub-list

Question

li = ['a', '#b', 'c']
repl = ['1', '2', '3']

I want to change '#b' with the repl sublist, meaning, this is my desired output:

['a', ['1', '2', '3'], 'c']

This is my attempt so far:

for i in li:
    if i.startswith('#'):
        i.replace(i, repl)

I get this error:

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-8-151b36fac37c> in <module>()
      1 for i in li:
      2     if i.startswith('#'):
----> 3         i.replace(i, repl)

TypeError: Can't convert 'list' object to str implicitly

Answer 1

You cannot replace a substring with a list. Instead, iterate over the list using a comprehension:

li = ['a', '#b', 'c']
repl = ['1', '2', '3']
new_li = [repl if i.startswith('#') else i for i in li]

Output:

['a', ['1', '2', '3'], 'c']

If you do not want to use list comprehension, you can either write out the entire generic loop, or use a functional approach:

Generic loop:

new_l = []
for i in li:
   if i.startswith('#'):
      new_l.append(repl)
   else:
      new_l.append(i)

Functional approach:

new_li = map(lambda x:repl if x.startswith('#') else x, li)

Answer 2

inplace version

li = ['a', '#b', 'c']
repl = ['1', '2', '3']
for i, item in enumerate(li):
    if item.startswith('#'):
        li[i] = repl

Answer 3

If you don't want to use list comprehension:

li = ['a', '#b', 'c']
repl = ['1', '2', '3']

for count, i in enumerate(li):
    if i.startswith('#'):
        li[count] = repl

print li

['a', ['1', '2', '3'], 'c']

Answer 4

li = ['a', '#b', 'c']
repl = ['1', '2', '3']

x = li.index('#b')
li[x] = repl
print(li)

this should give the desired result

Answer 5

You could also use a stack based approach:

li = ['a', '#b', 'c']
repl = ['1', '2', '3']

stack = []
for elem in li:
    stack.append(elem)
    if stack[-1].startswith("#"):
        stack.pop()
        stack.append(repl)

print(stack)

Which Outputs:

['a', ['1', '2', '3'], 'c']