CODEWITHSUNDEEP
pythonnumpy
DrApe
DrApe

Reputation: 131

Number of unique elements per row in a NumPy array

For example, for

a = np.array([[1, 0, 0], [1, 0, 0], [2, 3, 4]])

I want to get 

[2, 2, 3]

Is there a way to do this without for loops or using np.vectorize?

Edit: Actual data consists of 1000 rows of 100 elements each, with each element ranging from 1 to 365. The ultimate goal is to determine the percentage of rows that have duplicates. This was a homework problem which I already solved (with a for loop), but I was just wondering if there was a better way to do it with numpy.

Upvotes: 12

Views: 8401

Answers (4)

Divakar
Divakar

Reputation: 221574

Approach #1

One vectorized approach with sorting -

In [8]: b = np.sort(a,axis=1)

In [9]: (b[:,1:] != b[:,:-1]).sum(axis=1)+1
Out[9]: array([2, 2, 3])

Approach #2

Another method for ints that aren't very large would be with offsetting each row by an offset that would differentiate elements off each row from others and then doing binned-summation and counting number of non-zero bins per row -

n = a.max()+1
a_off = a+(np.arange(a.shape[0])[:,None])*n
M = a.shape[0]*n
out = (np.bincount(a_off.ravel(), minlength=M).reshape(-1,n)!=0).sum(1)

Runtime test

Approaches as funcs -

def sorting(a):
    b = np.sort(a,axis=1)
    return (b[:,1:] != b[:,:-1]).sum(axis=1)+1

def bincount(a):
    n = a.max()+1
    a_off = a+(np.arange(a.shape[0])[:,None])*n
    M = a.shape[0]*n
    return (np.bincount(a_off.ravel(), minlength=M).reshape(-1,n)!=0).sum(1)

# From @wim's post   
def pandas(a):
    df = pd.DataFrame(a.T)
    return df.nunique()

# @jp_data_analysis's soln
def numpy_apply(a):
    return np.apply_along_axis(compose(len, np.unique), 1, a)

Case #1 : Square shaped one

In [164]: np.random.seed(0)

In [165]: a = np.random.randint(0,5,(10000,10000))

In [166]: %timeit numpy_apply(a)
     ...: %timeit sorting(a)
     ...: %timeit bincount(a)
     ...: %timeit pandas(a)
1 loop, best of 3: 1.82 s per loop
1 loop, best of 3: 1.93 s per loop
1 loop, best of 3: 354 ms per loop
1 loop, best of 3: 879 ms per loop

Case #2 : Large number of rows

In [167]: np.random.seed(0)

In [168]: a = np.random.randint(0,5,(1000000,10))

In [169]: %timeit numpy_apply(a)
     ...: %timeit sorting(a)
     ...: %timeit bincount(a)
     ...: %timeit pandas(a)
1 loop, best of 3: 8.42 s per loop
10 loops, best of 3: 153 ms per loop
10 loops, best of 3: 66.8 ms per loop
1 loop, best of 3: 53.6 s per loop

Extending to number of unique elements per column

To extend, we just need to do the slicing and ufunc operations along the other axis for the two proposed approaches, like so -

def nunique_percol_sort(a):
    b = np.sort(a,axis=0)
    return (b[1:] != b[:-1]).sum(axis=0)+1

def nunique_percol_bincount(a):
    n = a.max()+1
    a_off = a+(np.arange(a.shape[1]))*n
    M = a.shape[1]*n
    return (np.bincount(a_off.ravel(), minlength=M).reshape(-1,n)!=0).sum(1)

Generic ndarray with generic axis

Let's see how we can extend to ndarray of generic dimensions and get those number of unique counts along a generic axis. We will make use of np.diff with its axis param to get those consecutive differences and hence make it generic, like so -

def nunique(a, axis):
    return (np.diff(np.sort(a,axis=axis),axis=axis)!=0).sum(axis=axis)+1

Sample runs -

In [77]: a
Out[77]: 
array([[1, 0, 2, 2, 0],
       [1, 0, 1, 2, 0],
       [0, 0, 0, 0, 2],
       [1, 2, 1, 0, 1],
       [2, 0, 1, 0, 0]])

In [78]: nunique(a, axis=0)
Out[78]: array([3, 2, 3, 2, 3])

In [79]: nunique(a, axis=1)
Out[79]: array([3, 3, 2, 3, 3])

If you are working with floating pt numbers and want to make the unique-ness case based on some tolerance value rather than absolute match, we can use np.isclose. Two such options would be -

(~np.isclose(np.diff(np.sort(a,axis=axis),axis=axis),0)).sum(axis)+1
a.shape[axis]-np.isclose(np.diff(np.sort(a,axis=axis),axis=axis),0).sum(axis)

For a custom tolerance value, feed those with np.isclose.

Upvotes: 13

jpp
jpp

Reputation: 164673

This solution via np.apply_along_axis isn't vectorised and involves a Python-level loop. But it is relatively intuitive using len + np.unique functions.

import numpy as np
from toolz import compose

a = np.array([[1, 0, 0], [1, 0, 0], [2, 3, 4]])

np.apply_along_axis(compose(len, np.unique), 1, a)    # [2, 2, 3]

Upvotes: 5

kuppern87
kuppern87

Reputation: 1135

A oneliner using sort:

In [6]: np.count_nonzero(np.diff(np.sort(a)), axis=1)+1
Out[6]: array([2, 2, 3])

Upvotes: 4

wim
wim

Reputation: 362716

Are you open to considering pandas? Dataframes have a dedicated method for this

>>> a = np.array([[1, 0, 0], [1, 0, 0], [2, 3, 4]])
>>> df = pd.DataFrame(a.T)
>>> print(*df.nunique())
2 2 3

Upvotes: 3

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