Replace the elements in list of lists according to the indexes stored in another list

Question

I have the following arrays:

example_array = [[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0]]
indexes = [[1,2,3], [2,3], [0,4,2], [1,4,3], [0,4,3], [1,2,3], [1,2]]

I would like to replace the elements in the example_array with 1 according to the index in indexes. So, the expected result should be this:

example_array = [[0,1,1,1,0],[0,0,1,1,0],[1,0,1,0,1],[0,1,0,1,1],[1,0,0,1,1],[0,1,1,1,0],[0,1,1,0,0]].

I tried different approaches, for example to use pandas, like this:

matrix = pd.DataFrame(example_array)

for row in matrix:
    for i in indexes:
        for j in i:
            x_u.iloc[x, j] = 1

but it doesn't give me the hoped results! The solution does not need to use pandas library. Thanks for the help!

pushkin · Accepted Answer · 2018-05-27 11:54:41Z

2

You can just do this:

example_array = [[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0]]
indexes = [[1,2,3], [2,3], [0,4,2], [1,4,3], [0,4,3], [1,2,3], [1,2]]

for i,index in enumerate(indexes):
  for idx in index:
    example_array[i][idx] = 1
print example_array

Output:

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

answered May 27, 2018 at 11:54

pushkin

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twolffpiggott · Accepted Answer · 2018-05-27 11:57:36Z

1

Try this:

example_array = [[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0]]
matrix = pd.DataFrame(example_array)
indexes = [[1,2,3], [2,3], [0,4,2], [1,4,3], [0,4,3], [1,2,3], [1,2]]
for row_num in range(len(matrix)):
    matrix.iloc[row_num][indexes[row_num]] = 1

It also saves you the nested loop!

answered May 27, 2018 at 11:57

twolffpiggott

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Asterisk · Accepted Answer · 2018-05-27 11:59:42Z

1

Using numpy array you can directly set specified indices to a desired value.

import numpy as np
example_array = np.array([[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0]])
indexes = [[1,2,3], [2,3], [0,4,2], [1,4,3], [0,4,3], [1,2,3], [1,2]]

for i, ind in enumerate(indexes):
    example_array[i, ind] = 1
print(a)

Output

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

answered May 27, 2018 at 11:59

Asterisk

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jamomani · Accepted Answer · 2018-05-27 12:05:12Z

1

example_array = [[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0]];
indexes = [[1,2,3], [2,3], [0,4,2], [1,4,3], [0,4,3], [1,2,3], [1,2]];

for i in range(0, len(example_array)):
    for j in range(0, len(indexes[i])):
        example_array[i][indexes[i][j]] = 1;

print example_array;

answered May 27, 2018 at 12:05

jamomani

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zipa · Accepted Answer · 2018-05-27 12:09:36Z

1

Or you can use the comprehension:

result = [[1 if i in ind else e for i, e in enumerate(ea)] for ea, ind in zip(example_array, indexes)]

answered May 27, 2018 at 12:09

zipa

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Reblochon Masque · Accepted Answer · 2018-05-27 12:09:36Z

1

with a one liner:

indices = [[1, 2, 3], [2, 3], [0, 4, 2], [1, 4, 3], [0, 4, 3], [1, 2, 3], [1, 2]]
print([[1 if idx in sub else 0 for idx in range(5)] for sub in indices])

output:

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

answered May 27, 2018 at 12:09

Reblochon Masque

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