Replace string by integer in sublists (all ocurrences) - python

Question

This an example of a possible list with the sublists:

  rows = [[1, 1, 'x'], [1, 'x', 1], [1, 1, 'x'], [1, 'x', 1], ['x', 1, 1], ['x', 1, 1]]

I am looking for an efficient way to replace 'x' either by 0 or 1 in the sublists. The sublists has no fixed sized or number. I mean, it could be a list like this one:

    [(1, 1, 'x', 'x'), (1, 1, 'x', 'x'), (1, 'x', 1, 'x'), (1, 'x', 'x', 1)]

I tried this:

  for row in rows:
      for pos in row:
        if pos == 'x':
           pos.replace('x','0')

Even knowning that I need '0' as an integer, just to print an output. But rows kept the 'x' unchanged.

Any suggestions?

Strings are immutable, replace returns a new string

Sayse
– Sayse

2019-10-08 12:46:15 +00:00
Commented Oct 8, 2019 at 12:46 — Sayse
– Sayse, Commented Oct 8, 2019 at 12:46

han solo · Accepted Answer · 2019-10-08 12:45:13Z

2

You could just use a list comprehension and check the value and replace that like,

>> rows = [[1, 1, 'x'], [1, 'x', 1], [1, 1, 'x'], [1, 'x', 1], ['x', 1, 1], ['x', 1, 1]]
>>> [[0 if y == 'x' else y  for y in x] for x in rows] # replace `x` with 0
[[1, 1, 0], [1, 0, 1], [1, 1, 0], [1, 0, 1], [0, 1, 1], [0, 1, 1]]

>>> another = [(1, 1, 'x', 'x'), (1, 1, 'x', 'x'), (1, 'x', 1, 'x'), (1, 'x', 'x', 1)]
>>> [[5 if y == 'x' else y  for y in x] for x in another] # replace `x` with 5
[[1, 1, 5, 5], [1, 1, 5, 5], [1, 5, 1, 5], [1, 5, 5, 1]]

answered Oct 8, 2019 at 12:45

han solo

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Comments

Amit Yadav · Accepted Answer · 2019-10-08 14:42:32Z

2

Try this code:

rows = [[1, 1, 'x'], [1, 'x', 1], [1, 1, 'x'], [1, 'x', 1], ['x', 1, 1], ['x', 1, 1]]

temp_dic = {'x':0}
modified_list  = map(
  lambda sub_list: sub_list if 'x' not in sub_list else [temp_dic.get(n, n) for n in sub_list],
  rows)
print(list(modified_list))

Output:

[[1, 1, 0], [1, 0, 1], [1, 1, 0], [1, 0, 1], [0, 1, 1], [0, 1, 1]]

Basically, we are iterating through the sublist and checking if a value (0) is present for the key (x). If yes, replace it, otherwise, keep the key.

dic.get(n, n) returns the value(0) if it exists or it simply just returns key (e.g numbers in our case). This way we only replace the 'x' not the numbers.

edited Oct 8, 2019 at 14:42

answered Oct 8, 2019 at 12:58

Amit Yadav

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2 Comments

powerPixie Over a year ago

When I run the code I got this: NameError: name 'dic' is not defined

Amit Yadav Over a year ago

Fixed it! It was a variable name typo. try now

Teku1998 · Accepted Answer · 2019-10-08 13:07:14Z

1

rows = [[1, 1, 'x'], [1, 'x', 1], [1, 1, 'x'], [1, 'x', 1], ['x', 1, 1], ['x', 1, 1]]

for r in rows:
    for i, pos in enumerate(r):
        if pos == 'x':
            r[i] = 0
print(rows)

#output replace x with 0: [[1, 1, 0], [1, 0, 1], [1, 1, 0], [1, 0, 1], [0, 1, 1], [0, 1, 1]]
#output replace x with 99: [[1, 1, 99], [1, 99, 1], [1, 1, 99], [1, 99, 1], [99, 1, 1], [99, 1, 1]]

This works well for me on Python 3.7.4. Cheers and good luck!

answered Oct 8, 2019 at 13:07

Teku1998

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Comments

vendrediSurMer · Accepted Answer · 2019-10-08 13:27:03Z

0

If you still wanted to use the replace method, you would need to append the values to a new list after replacing the string:

rows=[(1, 1, 'x', 'x'), (1, 1, 'x', 'x'), (1, 'x', 1, 'x'), (1, 'x', 'x', 1)]

new_list=[]
for row in rows:
    sublist=[]
    for pos in row:
        if pos == 'x':
            pos=pos.replace('x','0')
        sublist.append(pos)
    new_list.append(sublist)

print(new_list)
# [[1, 1, '0', '0'], [1, 1, '0', '0'], [1, '0', 1, '0'], [1, '0', '0', 1]]

answered Oct 8, 2019 at 13:27

vendrediSurMer

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1 Comment

powerPixie Over a year ago

As I said, I need to keep the '0's or '1's as integers. Your solution is not good for me in this particular case. Thanks, anyway.

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