Python numpy array: wrong result when mixing int32 and int8

Question

I saw a very strange behavior in numpy array, when I mixed int32 and int8 arrays in a simple operation, the int32 array element ct[4,0] seems to have become 8bit when taking the result of += dleng[4]*4:

import numpy as np
In[3]: ct = np.zeros((6,1), np.int32)
In[4]: ct
Out[4]: 
array([[0],
   [0],
   [0],
   [0],
   [0],
   [0]], dtype=int32)
In[5]: dleng = np.zeros((6, 1), np.int8)
In[6]: dleng[0] = 2
dleng[1] = 3
dleng[2] = 4
dleng[3] = 7
dleng[4] = 3
dleng[5] = 5
In[7]: dleng
Out[7]: 
array([[2],
   [3],
   [4],
   [7],
   [3],
   [5]], dtype=int8)
In[8]: ct[4] = 117
In[9]: ct
Out[9]: 
array([[  0],
   [  0],
   [  0],
   [  0],
   [117],
   [  0]], dtype=int32)
In[10]: ct[4,0] += dleng[4]*4
In[11]: ct
Out[11]: 
array([[   0],
   [   0],
   [   0],
   [   0],
   [-127],
   [   0]], dtype=int32)}

Does anyone know why this might happen?

Warren Weckesser · Accepted Answer · 2015-01-30 20:28:20Z

1

dleng[4]*4 is an array:

In [94]: dleng[4]
Out[94]: array([3], dtype=int8)

In [95]: dleng[4]*4
Out[95]: array([12], dtype=int8)

while ct[4, 0] is a scalar with type np.int32:

In [98]: ct[4,0]
Out[98]: 117

In [99]: type(_)
Out[99]: numpy.int32

As @WallyBeaver points out, ct[4,0] += dleng[4]*4 is like ct[4,0] = ct[4,0] + dleng[4]*4. That last expression is a scalar plus an array. In that situation, the data type is determined by the array, so it ends up being np.int8. There is a note about this in the numpy docs:

Mixed scalar-array operations use a different set of casting rules that ensure that a scalar cannot “upcast” an array unless the scalar is of a fundamentally different kind of data (i.e., under a different hierarchy in the data-type hierarchy) than the array. This rule enables you to use scalar constants in your code (which, as Python types, are interpreted accordingly in ufuncs) without worrying about whether the precision of the scalar constant will cause upcasting on your large (small precision) array.

A fix is to write the in-place addition as

ct[4,0] += dleng[4,0]*4

answered Jan 30, 2015 at 20:28

Warren Weckesser

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1 Comment

fpeng Over a year ago

Ok this makes sense. Definitely something to pay attention to in the future!

user4482921 · Accepted Answer · 2015-01-30 20:13:48Z

1

It's because what you are doing is:

>>> ct[4,0] += dleng[4]*4

Which is actually this:

>>> ct[4,0] = ct[4,0] + dleng[4]*4

Which produces this:

>>> ct[4,0] + dleng[4]*4
array([-127], dtype=int8)

Which then does this:

>>> ct[4,0] = array([-127], dtype=int8)

But is really doing this type conversion under the covers:

>>> a.astype(np.int32)
array([-127], dtype=int32)

answered Jan 30, 2015 at 20:13

user4482921

3 Comments

fpeng Over a year ago

I don't understand why ct[4,0] + dleng[4]*4 produces array([-127], dtype=int8) --- why is it cast to int8 instead of int32?

fpeng Over a year ago

Ah I understand now after seeing both yours and Warren's answers.

user4482921 Over a year ago

Yeah, ct[4,0] is 117, which is then cast to an int8 upon addition.

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