performance - Efficiently replace elements in array based on dictionary - NumPy / Python

Question

First, of all, my apologies if this has been answered elsewhere. All I could find were questions about replacing elements of a given value, not elements of multiple values.

background

I have several thousand large np.arrays, like so:

# generate dummy data
input_array = np.zeros((100,100))
input_array[0:10,0:10] = 1
input_array[20:56, 21:43] = 5
input_array[34:43, 70:89] = 8

In those arrays, I want to replace values, based on a dictionary:

mapping = {1:2, 5:3, 8:6}

approach

At this time, I am using a simple loop, combined with fancy indexing:

output_array = np.zeros_like(input_array)

for key in mapping:
    output_array[input_array==key] = mapping[key]

problem

My arrays have dimensions of 2000 by 2000, the dictionaries have around 1000 entries, so, these loops take forever.

question

is there a function, that simply takes an array and a mapping in the form of a dictionary (or similar), and outputs the changed values?

help is greatly appreciated!

Update:

Solutions:

I tested the individual solutions in Ipython, using

%%timeit -r 10 -n 10

input data

import numpy as np
np.random.seed(123)

sources = range(100)
outs = [a for a in range(100)]
np.random.shuffle(outs)
mapping = {sources[a]:outs[a] for a in(range(len(sources)))}

For every solution:

np.random.seed(123)
input_array = np.random.randint(0,100, (1000,1000))

divakar, method 3:

%%timeit -r 10 -n 10
k = np.array(list(mapping.keys()))
v = np.array(list(mapping.values()))

mapping_ar = np.zeros(k.max()+1,dtype=v.dtype) #k,v from approach #1
mapping_ar[k] = v
out = mapping_ar[input_array]

5.01 ms ± 641 μs per loop (mean ± std. dev. of 10 runs, 10 loops each)

divakar, method 2:

%%timeit -r 10 -n 10
k = np.array(list(mapping.keys()))
v = np.array(list(mapping.values()))

sidx = k.argsort() #k,v from approach #1

k = k[sidx]
v = v[sidx]

idx = np.searchsorted(k,input_array.ravel()).reshape(input_array.shape)
idx[idx==len(k)] = 0
mask = k[idx] == input_array
out = np.where(mask, v[idx], 0)

56.9 ms ± 609 μs per loop (mean ± std. dev. of 10 runs, 10 loops each)

divakar, method 1:

%%timeit -r 10 -n 10

k = np.array(list(mapping.keys()))
v = np.array(list(mapping.values()))

out = np.zeros_like(input_array)
for key,val in zip(k,v):
    out[input_array==key] = val

113 ms ± 6.2 ms per loop (mean ± std. dev. of 10 runs, 10 loops each)

eelco:

%%timeit -r 10 -n 10
output_array = npi.remap(input_array.flatten(), list(mapping.keys()), list(mapping.values())).reshape(input_array.shape)

143 ms ± 4.47 ms per loop (mean ± std. dev. of 10 runs, 10 loops each)

yatu

%%timeit -r 10 -n 10

keys, choices = list(zip(*mapping.items()))
# [(1, 5, 8), (2, 3, 6)]
conds = np.array(keys)[:,None,None]  == input_array
np.select(conds, choices)

157 ms ± 5 ms per loop (mean ± std. dev. of 10 runs, 10 loops each)

original, loopy method:

%%timeit -r 10 -n 10
output_array = np.zeros_like(input_array)

for key in mapping:
    output_array[input_array==key] = mapping[key]

187 ms ± 6.44 ms per loop (mean ± std. dev. of 10 runs, 10 loops each)

Thanks for the superquick help!

Answer 1

Approach #1 : Loopy one with array data

One approach would be extracting the keys and values in arrays and then use a similar loop -

k = np.array(list(mapping.keys()))
v = np.array(list(mapping.values()))

out = np.zeros_like(input_array)
for key,val in zip(k,v):
    out[input_array==key] = val

Benefit with this one over the original one is the spatial-locality of the array data for efficient data-fetching, which is used in the iterations.

Also, since you mentioned thousand large np.arrays. So, if the mapping dictionary stays the same, that step to get the array versions - k and v would be a one-time setup process.

Approach #2 : Vectorized one with searchsorted

A vectorized one could be suggested using np.searchsorted -

sidx = k.argsort() #k,v from approach #1

k = k[sidx]
v = v[sidx]

idx = np.searchsorted(k,input_array.ravel()).reshape(input_array.shape)
idx[idx==len(k)] = 0
mask = k[idx] == input_array
out = np.where(mask, v[idx], 0)

Approach #3 : Vectorized one with mapping-array for integer keys

A vectorized one could be suggested using a mapping array for integer keys, which when indexed by the input array would lead us directly to the final output -

mapping_ar = np.zeros(k.max()+1,dtype=v.dtype) #k,v from approach #1
mapping_ar[k] = v
out = mapping_ar[input_array]