Python - Error when i replace data of an array by 2 values of a second array

Question

I have two numpy arrays "Elements" and "nodes". My aim is to gather some data of these arrays. I need to remplace "Elements" data of the two last columns by the two coordinates contains in "nodes" array. The two arrays are very huge, i have to automate it.

An example :

import numpy as np

Elements = np.array([[1.,11.,14.],[2.,12.,13.]])

nodes = np.array([[11.,0.,0.],[12.,1.,1.],[13.,2.,2.],[14.,3.,3.]])

results = np.array([[1., 0., 0., 3., 3.],
[2., 1., 1., 2., 2.]])

On a last post Stack', someone helps me to do it :

e = Elements[:,1:].ravel().astype(int)
n=nodes[:,0].astype(int)
I, J = np.where(e==n[:,None])
results = np.zeros((e.shape[0],2),nodes.dtype)
results[J] = nodes[I,1:]
np.concatenate((Elements[:,[0]],results.reshape(2,4)),axis=1)

It works here, but when i do on a great number of values, i got an error : "need more than 1 value to unpack" but i dont understand what is the origin of the problem...

--- EDIT LATER -----

I think it could be due to the importance of my arrays. Perhaps is there an other way to deal with this problem? (21536, 4) and Nodes_coord.shape : (10926, 3)

On the " I, J = np.where(e==n[:,None])" I dont know if my values present something wrong... — user3601754
– user3601754, Commented Jan 4, 2016 at 10:20
In the occasion that you get an error, I'm guessing that e==n[:,None] is never true, in which case np.where returns a tuple with only a single value, which you're trying to unpack into two values (I and J). — acdr
– acdr, Commented Jan 4, 2016 at 10:24
Table_connect.shape : (21536, 4) and Nodes_coord.shape : (10926, 3) But it is impossible to be never True :/ A strange value could be present in my data which provoks it? — user3601754
– user3601754, Commented Jan 4, 2016 at 10:24

acdr · Accepted Answer · 2016-01-04 10:41:40Z

1

It looks like you're using at least one element which does not have a corresponding node. Try this code to check if this is indeed the case:

element_ids = Elements[:, 1]
node_ids = nodes[:, 0]
valid = np.all([elem_id in node_ids for elem_id in element_ids ])

I'm guessing valid will be False.

answered Jan 4, 2016 at 10:41

acdr

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5 Comments

user3601754 Over a year ago

Thanks for help and I get "True" :/

acdr Over a year ago

In that case, what does np.where(e==n[:,None]) evaluate to?

user3601754 Over a year ago

It returns (array([], dtype=int32),) very strange :/

acdr Over a year ago

This implies that elements in e are never equal to elements in n[:, None], which implies that your elements do not have matching nodes.

user3601754 Over a year ago

Yep but it seems to be impossible :/ thanks for your help ;)

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