changing values of an array

Question

I have a given numpy array, data. I want to change its values in the ascending order from 0. I could solve it by manually making the look up table, LUT.

import numpy as np

data = np.array([[2, 6, 16, 39, 43],
                        [43, 6, 16, 39, 2]])


LUT={2:0,
    6:1,
    16:2,
    39:3,
    43:4} 

def changer(X):   
    res = np.copy(X)    
    for i, j in LUT.items():
        res[X==i] = j
    return res       

result = changer(data)
print (result)

[[0 1 2 3 4]
 [4 1 2 3 0]]

The result is correct as I expected. However, sometimes, I am lazy to make LUT manually. So, how could I get the same results programmatically?

edit

I tried to make dictionary, LUT as follows.

list = [2,6,16,39,43]

LUT = {}
for i in enumerate(list):
    LUT.update({list[i]: i})

But, 

D.update({list[i]: i})
TypeError: list indices must be integers, not tuple

It is not a good idea to hard code any look up tables in high level code into python! Except with some edge case scenarios this will hurt you. Please explain what you are trying to do and why are you trying to do so. — not_a_bot_no_really_82353, Dec 10 '18 at 18:46
Possible duplicate of [Rank items in an array using Python/NumPy](https://stackoverflow.com/questions/5284646/rank-items-in-an-array-using-python-numpy) — Ben.T, Dec 10 '18 at 18:48
Are you looking for something like this?`dict(enumerate([2, 6, 16, 39, 43]))` — mad_, Dec 10 '18 at 18:49
@mad_ how can i inverse, dict(enumerate([2, 6, 16, 39, 43]))? — , Dec 10 '18 at 18:54
By converting values to keys `dict(zip(d.values(),d.keys()))` — mad_, Dec 10 '18 at 18:55

score 0 · Answer 1 · answered Dec 10 '18 at 21:45

Use np.unique with the return_inverse flag:

data = np.array([[2, 6, 16, 39, 43],
                 [43, 6, 16, 39, 2]])

unq, ranked = np.unique(data, return_inverse=True)

unique without axis set flattens, so we need to reshape

ranked = ranked.reshape(data.shape)
ranked
# array([[0, 1, 2, 3, 4],
#        [4, 1, 2, 3, 0]])

score 0 · Answer 2 · answered Dec 10 '18 at 21:53

If I understand what you want, np_lu should be automatically generated from your data.

In [24]: np_lu=sorted(set(data.ravel()))
In [25]: np_lu
Out[25]: [2, 6, 16, 39, 43]

Create an array of indices 2->0, 6->1, ... 43->4

In [29]: indx=np.array(np_lu)
In [33]: d=np.zeros(indx.max()+1)
In [34]: d[indx]=np.arange(len(indx))
In [35]: d
Out[35]:
array([ 0.,  0.,  0.,  0.,  0.,  0.,  1.,  0.,  0.,  0.,  0.,  0.,  0.,
    0.,  0.,  0.,  2.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,
    0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,
    3.,  0.,  0.,  0.,  4.])

In [36]: data
Out[36]:
array([[ 2,  6, 16, 39, 43],
       [43,  6, 16, 39,  2]])

Create a result array, res.

In [37]: res=np.zeros_like(data)

Fill res with the data in d, indexed by data.

In [41]: res=d[data]
In [42]: res
Out[42]:
array([[ 0.,  1.,  2.,  3.,  4.],
       [ 4.,  1.,  2.,  3.,  0.]])

This will only work for data of int type. There's probably a lot of tidying up that can be done in my code.

Scott Boston · Answer 3 · 2018-12-12T14:19:11.027

0

I think this is really `np.argsort`:

Returns the indices that would sort an array.

import numpy as np
data = np.array([[2, 6, 16, 39, 43],
                        [43, 6, 16, 39, 2]])
data.argsort(1).tolist()

Output:

[[0, 1, 2, 3, 4], [4, 1, 2, 3, 0]]

And to go backwards giving 1D array, l, and 2D array of indexes, iarr:

l = np.array([2, 6, 16, 39, 43])
iarr = np.array([[0, 1, 2, 3, 4], [4, 1, 2, 3, 0]])

We can use broadcasting:

l[iarr]

Output:

array([[ 2,  6, 16, 39, 43],
       [43,  6, 16, 39,  2]])

edited Dec 12 '18 at 14:19

answered Dec 10 '18 at 21:55

Scott Boston

147,308
15
139
187

Is there also a way to reverse the output to input? – Dec 11 '18 at 06:03
Reverse the output? Change the sort order? [[4->0],[0->4]] like this? – Scott Boston Dec 11 '18 at 12:04
to get `input` from `output`. – Dec 12 '18 at 05:39
`[[0, 1, 2, 3, 4], [4, 1, 2, 3, 0]]` into `[[2, 6, 16, 39, 43], [43, 6, 16, 39, 2]]` – Dec 12 '18 at 05:40
Given a list of [2,6,16,39,43] and [[0, 1, 2, 3, 4], [4, 1, 2, 3, 0]] -> [[2, 6, 16, 39, 43], [43, 6, 16, 39, 2]]? – Scott Boston Dec 12 '18 at 12:09
but we do not know `l`, we know only `data` and `iarr` – Dec 13 '18 at 08:06
Yes, then it is not possible. You can't got from a an index to numbers with a reference. – Scott Boston Dec 13 '18 at 13:33

masasa · Answer 4 · 2018-12-10T19:06:52.297

-1

if i get right your needs , all you need is:

import numpy as np
lst = np.array([[i for i in range(len(data[0]))],[i for i in range(len(data[1])-1,-1,-1)]])

edited Dec 10 '18 at 19:06

answered Dec 10 '18 at 19:01

masasa

260
1
9

Nope. Not even close. OP is looking for functionality similar to lookup. Your code just iterates and print index of items in some fashion. – mad_ Dec 10 '18 at 19:17
he wanted to use lookup as a method to resolve his problems, looking over his desired outcome , i think this nails it... – masasa Dec 10 '18 at 19:18
Can you confirm if the output matches? This does not give the desired results. Thanks – mad_ Dec 10 '18 at 19:22
@mad_ i ran it on his initial data set (array with 2 lists)with my commands and it worked – masasa Dec 10 '18 at 19:26

changing values of an array

edit

4 Answers4

I think this is really np.argsort:

And to go backwards giving 1D array, l, and 2D array of indexes, iarr:

I think this is really `np.argsort`: