Numpy ValueError: setting an array element with a sequence. This message may appear without the existing of a sequence?

Question:

Why do I get this error message? ValueError: setting an array element with a sequence. Thank you

Z=np.array([1.0,1.0,1.0,1.0])  

def func(TempLake,Z):
    A=TempLake
    B=Z
    return A*B

Nlayers=Z.size
N=3
TempLake=np.zeros((N+1,Nlayers))

kOUT=np.zeros(N+1)
for i in xrange(N):
    kOUT[i]=func(TempLake[i],Z)
Asked By: user1419224

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Answers:

You’re getting the error message

ValueError: setting an array element with a sequence.

because you’re trying to set an array element with a sequence. I’m not trying to be cute, there — the error message is trying to tell you exactly what the problem is. Don’t think of it as a cryptic error, it’s simply a phrase. What line is giving the problem?

kOUT[i]=func(TempLake[i],Z)

This line tries to set the ith element of kOUT to whatever func(TempLAke[i], Z) returns. Looking at the i=0 case:

In [39]: kOUT[0]
Out[39]: 0.0

In [40]: func(TempLake[0], Z)
Out[40]: array([ 0.,  0.,  0.,  0.])

You’re trying to load a 4-element array into kOUT[0] which only has a float. Hence, you’re trying to set an array element (the left hand side, kOUT[i]) with a sequence (the right hand side, func(TempLake[i], Z)).

Probably func isn’t doing what you want, but I’m not sure what you really wanted it to do (and don’t forget you can usually use vectorized operations like A*B rather than looping in numpy.) That should explain the problem, anyway.

Answered By: DSM

I believe python arrays just admit values. So convert it to list:

kOUT = np.zeros(N+1)
kOUT = kOUT.tolist()
Answered By: Luis Renato

It’s a pity that both of the answers analyze the problem but didn’t give a direct answer.
Let’s see the code.

Z = np.array([1.0, 1.0, 1.0, 1.0])  

def func(TempLake, Z):
    A = TempLake
    B = Z
    return A * B
Nlayers = Z.size
N = 3
TempLake = np.zeros((N+1, Nlayers))
kOUT = np.zeros(N + 1)

for i in xrange(N):
    # store the i-th result of
    # function "func" in i-th item in kOUT
    kOUT[i] = func(TempLake[i], Z)

The error shows that you set the ith item of kOUT(dtype:int) into an array. Here every item in kOUT is an int, can’t directly assign to another datatype. Hence you should declare the data type of kOUT when you create it.
For example, like:

Change the statement below:

kOUT = np.zeros(N + 1)

into:

kOUT = np.zeros(N + 1, dtype=object)

or:

kOUT = np.zeros((N + 1, N + 1))

All code:

import numpy as np
Z = np.array([1.0, 1.0, 1.0, 1.0])

def func(TempLake, Z):
    A = TempLake
    B = Z
    return A * B

Nlayers = Z.size
N = 3
TempLake = np.zeros((N + 1, Nlayers))

kOUT = np.zeros(N + 1, dtype=object)
for i in xrange(N):
    kOUT[i] = func(TempLake[i], Z)

Hope it can help you.

Answered By: Junning Huang

KOUT[i] is a single element of a list. But you are assigning a list to this element. your func is generating a list.

Answered By: Shahzamal
Z=np.array([1.0,1.0,1.0,1.0])  

def func(TempLake,Z):
    A=TempLake
    B=Z
    return A*B
Nlayers=Z.size
N=3
TempLake=np.zeros((N+1,Nlayers))
kOUT=np.vectorize(func)(TempLake,Z)

This works too , instead of looping , just vectorize however read below notes from the scipy documentation : https://docs.scipy.org/doc/numpy/reference/generated/numpy.vectorize.html

The vectorize function is provided primarily for convenience, not for performance. The implementation is essentially a for loop.

If otypes is not specified, then a call to the function with the first argument will be used to determine the number of outputs. The results of this call will be cached if cache is True to prevent calling the function twice. However, to implement the cache, the original function must be wrapped which will slow down subsequent calls, so only do this if your function is expensive.

Answered By: Ahalya L

To put a sequence or another numpy array into a numpy array,
Just change this line:

kOUT=np.zeros(N+1)

to:

kOUT=np.asarray([None]*(N+1))

Or:

kOUT=np.zeros((N+1), object)
Answered By: Oluwasegun Somefun

You can try the expand option in Series.str.split('seperator', expand=True).
By default expand is False.

expand : bool, default False
Expand the splitted strings into separate columns.

  • If True, return DataFrame/MultiIndex expanding dimensionality.
  • If False, return Series/Index, containing lists of strings.
Answered By: Kenny
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