How to grab corresponding values from two-dimensional numpy array

Question

I have a two-dimension array

[[ 1,  4,  4,  6,  6,  6,  8,  8,  8,  8,  9,  9,  9,  9,  9, 10, 11, 11, 11, 11, 11], 
[ 0,  0,  1,  0,  1,  4,  0,  1,  4,  6,  0,  1,  4,  6,  8,  5,  0,  1,  4,  6,  8]]

I'm trying to parse it in such a manner that I end up with output that looks like this. Doesn't have to be dictionary but for visualizing I'm using something of the sort. Find every second coord for every unique value in the first array.

{
1: [0],
4: [0,1],
6: [0,1,4],
8: [0,1,4,6],
9: [0,1,4,6,8]
}

1 only had one second coord, of 0 4 had matching coords of 0 and 1 etc.

Any thoughts?

Thanks!

Corralien · Accepted Answer · 2022-06-28 21:56:40Z

Update

Using Python:

from itertools import groupby

lst = [[1, 4, 4, 6, 6, 6, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 11, 11, 11, 11, 11],
       [0, 0, 1, 0, 1, 4, 0, 1, 4, 6, 0, 1, 4, 6, 8, 5, 0, 1, 4, 6, 8]]

# Dict version
out = {k: [v[1] for v in l] for k, l in groupby(zip(*lst), key=lambda x: x[0])}
print(out)

# Output
{1: [0],
 4: [0, 1],
 6: [0, 1, 4],
 8: [0, 1, 4, 6],
 9: [0, 1, 4, 6, 8],
 10: [5],
 11: [0, 1, 4, 6, 8]}

# List version
out = [[v[1] for v in l] for k, l in groupby(zip(*lst), key=lambda x: x[0])]

Old answer

Using Pandas:

lst = [[1, 4, 4, 6, 6, 6, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 11, 11, 11, 11, 11],
       [0, 0, 1, 0, 1, 4, 0, 1, 4, 6, 0, 1, 4, 6, 8, 5, 0, 1, 4, 6, 8]]

df = pd.DataFrame(lst).T

# Dict version
out = df.groupby(0)[1].agg(list).to_dict()
print(out)

# Output
{1: [0],
 4: [0, 1],
 6: [0, 1, 4],
 8: [0, 1, 4, 6],
 9: [0, 1, 4, 6, 8],
 10: [5],
 11: [0, 1, 4, 6, 8]}

# List version
out = df.groupby(0)[1].agg(list).to_list()

Sabil · Accepted Answer · 2022-06-28 21:46:17Z

1

The following code seems to meet your output preferences.

Code:

val_list = [
  [ 1,  4,  4,  6,  6,  6,  8,  8,  8,  8,  9,  9,  9,  9,  9, 10, 11, 11, 11, 11, 11],
  [ 0,  0,  1,  0,  1,  4,  0,  1,  4,  6,  0,  1,  4,  6,  8,  5,  0,  1,  4,  6,  8]
]

val_dict = {}
for i, j in zip(val_list[0], val_list[1]):
  if i in val_dict:
    val_dict[i].append(j)
  else:
    val_dict[i] = [j]

print(val_dict)

Output:

{1: [0], 4: [0, 1], 6: [0, 1, 4], 8: [0, 1, 4, 6], 9: [0, 1, 4, 6, 8], 10: [5], 11: [0, 1, 4, 6, 8]}

answered Jun 28, 2022 at 21:46

Sabil

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