Test of equality between arrays using in1d function

Question

Before I ask my question, I provide you with the code:

from scipy import *
x = randn(10)
cum_x = cumsum(x)
# The objective is to recover x using cum_x and the diff function.
y = append(cum_x[0], diff(cum_x))
# Now, y should be equal to x but this is not confirmed by the function in1d
test = in1d(x,y)

The variable test does not return an array of "True" boolean values even if y and x are clearly the same. What is the problem here?

I imagine there are floating point issues, allclose might be what you want — Padraic Cunningham
– Padraic Cunningham, Commented Jan 3, 2016 at 15:26
It is. But why in1d does not return an array of True values? Why rounding issues? Do you think the cumulative sum and the difference produce slightly different floats? — Charlie
– Charlie, Commented Jan 3, 2016 at 15:28
I added an answer showing increased precision, you can see there are slight differences — Padraic Cunningham
– Padraic Cunningham, Commented Jan 3, 2016 at 15:31
Aside: even if x and y weren't clearly different, in1d isn't a good way to check to see if they're equal: -- in1d([1,2],[2,1]) gives all true, after all. — DSM
– DSM, Commented Jan 3, 2016 at 15:31

smci · Accepted Answer · 2024-02-05 01:28:23Z

3

if you use set_printoptions to increase precision you will see some differences:

import numpy as np
from scipy import *
np.set_printoptions(30)
x = randn(10)
cum_x = cumsum(x)

#The objective is to recover x using cum_x and the diff function.
y = append(cum_x[0], diff(cum_x))
print(x)
print("\n")
print(y)
#Now, y should be equal to x but this is not confirmed by the function in1d
test = in1d(x, y)
print(test)

Output:

[ 0.54816314147543721002620031868   0.14319052613251953554041051575
  0.489110961092741158839913850898 -0.093011827554544138085823590245
 -0.58370623188476589149331630324  -0.40395493550429123486011917521
  0.387387395892057895263604905267  1.001637373359834937147638811439
 -1.486778459872974744726548124163  1.446772274227251076084144187917]


[ 0.54816314147543721002620031868   0.143190526132519591051561747008
  0.48911096109274110332876261964  -0.093011827554544179719187013688
 -0.58370623188476589149331630324  -0.40395493550429123486011917521
  0.387387395892057895263604905267  1.001637373359834937147638811439
 -1.486778459872974744726548124163  1.446772274227251076084144187917]
[ True False False False  True  True  True  True  True  True]

What you probably want is allclose but interestingly setting the dtype to np.float128 or np.longdouble on my ubuntu system does not lose precision and in1d returns True.

 cum_x = cumsum(x,dtype=np.longdouble)

edited Feb 5, 2024 at 1:28

smci

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answered Jan 3, 2016 at 15:28

Padraic Cunningham

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3 Comments

Charlie Over a year ago

So do you think the cumulative sum and the difference functions produce these slight differences?

Padraic Cunningham Over a year ago

@Charlie, if you look at the output they definitely do, what you want to be using is allclose.

smci Over a year ago

set_printoptions() is in numpy package, not scipy

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