I have an array of integers:
import numpy as np
demo = np.array([[1, 2, 3],
[1, 5, 3],
[4, 5, 6],
[7, 8, 9],
[4, 2, 3],
[4, 2, 12],
[10, 11, 13]])
And I want an array of unique values in the columns, padded with something if necessary (e.g. nan):
[[1, 4, 7, 10, nan],
[2, 5, 8, 11, nan],
[3, 6, 9, 12, 13]]
It does work when I iterate over the transposed array and use a boolean_indexing solution from a previous question. But I was hoping there would be a built-in method:
solution = []
for row in np.unique(demo.T, axis=1):
solution.append(np.unique(row))
def boolean_indexing(v, fillval=np.nan):
lens = np.array([len(item) for item in v])
mask = lens[:,None] > np.arange(lens.max())
out = np.full(mask.shape,fillval)
out[mask] = np.concatenate(v)
return out
print(boolean_indexing(solution))
AFAIK, there are no builtin solution for that. That being said, your solution seems a bit complex to me. You could create an array with initialized values and fill it with a simple loop (since you already use loops anyway).
solution = [np.unique(row) for row in np.unique(demo.T, axis=1)]
result = np.full((len(solution), max(map(len, solution))), np.nan)
for i,arr in enumerate(solution):
result[i][:len(arr)] = arr
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я говорю два предложения