# How to get a list of all indices of repeated elements in a numpy array

A vectorized solution with numpy, on the magic of `unique()`.

``````import numpy as np

# create a test array
records_array = np.array([1, 2, 3, 1, 1, 3, 4, 3, 2])

# creates an array of indices, sorted by unique element
idx_sort = np.argsort(records_array)

# sorts records array so all unique elements are together
sorted_records_array = records_array[idx_sort]

# returns the unique values, the index of the first occurrence of a value, and the count for each element
vals, idx_start, count = np.unique(sorted_records_array, return_counts=True, return_index=True)

# splits the indices into separate arrays
res = np.split(idx_sort, idx_start[1:])

#filter them with respect to their size, keeping only items occurring more than once
vals = vals[count > 1]
res = filter(lambda x: x.size > 1, res)
``````

The following code was the original answer, which required a bit more memory, using `numpy` broadcasting and calling `unique` twice:

``````records_array = array([1, 2, 3, 1, 1, 3, 4, 3, 2])
vals, inverse, count = unique(records_array, return_inverse=True,
return_counts=True)

idx_vals_repeated = where(count > 1)[0]
vals_repeated = vals[idx_vals_repeated]

rows, cols = where(inverse == idx_vals_repeated[:, newaxis])
_, inverse_rows = unique(rows, return_index=True)
res = split(cols, inverse_rows[1:])
``````

with as expected `res = [array([0, 3, 4]), array([1, 8]), array([2, 5, 7])]`