Insert a list, element by element, into the elements of an array of arrays

Question

I have an array of arrays:

parameters = [np.array([ 2.1e-04, -8.3e-03, 9.8e-01]), np.array([ 5.5e-04, 1.2e-01, 9.9e-01]), ...]

whose length is:

print len(parameters)

100

If we label the elements of parameters as parameters[i][j]:

it is then possible to access each number, i.e. print parameters[1][2] gives 0.99

I also have an array:

 temperatures = [110.51, 1618.079, ...]

whose length is also 100:

print len(temperatures)

100

Let the elements of temperatures be k:

I would like to insert each kth element of temperatures into each ith element of parameters, in order to obtain final:

final = [np.array([ 2.1e-04, -8.3e-03, 9.8e-01, 110.51]), np.array([ 5.5e-04, 1.2e-01, 9.9e-01, 1618.079]), ...]

I have tried to make something like a zip loop:

for i,j in zip(parameters, valid_temperatures): final = parameters[2][i].append(valid_temperatures[j])

but this does not work. I would appreciate if you could help me.

EDIT: Based on @hpaulj answer:

If you run Solution 1:

 parameters = [np.array([ 2.1e-04, -8.3e-03, 9.8e-01]), np.array([ 5.5e-04, 1.2e-01, 9.9e-01])]
 temperatures = [110.51, 1618.079]

     for i,(arr,t) in enumerate(zip(parameters,temperatures)):
     parameters[i] = np.append(arr,t)

 print parameters

It gives:

 [array([  2.10000000e-04,  -8.30000000e-03,   9.80000000e-01,
     1.10510000e+02]), array([  5.50000000e-04,   1.20000000e-01,   9.90000000e-01,
     1.61807900e+03])]

which is the desired output.

In addition, Solution 2:

parameters = [np.array([ 2.1e-04, -8.3e-03, 9.8e-01]), np.array([ 5.5e-04, 1.2e-01, 9.9e-01])]
temperatures = [110.51, 1618.079]

parameters = [np.append(arr,t) for arr, t in zip(parameters,temperatures)]

print parameters

also gives the desired output.

As opposed to Solution 1, Solution 2 doesn't use the ith enumerate index. Therefore, if I just split Solution 2's [np.append ... for arr ] syntax the following way:

 parameters = [np.array([ 2.1e-04, -8.3e-03, 9.8e-01]), np.array([ 5.5e-04, 1.2e-01, 9.9e-01])]
 temperatures = [110.51, 1618.079]

 for arr, t in zip(parameters,temperatures):
   parameters = np.append(arr,t)   

 print parameters

The output contains only the last iteration, and not in an "array-format":

 [  5.50000000e-04   1.20000000e-01   9.90000000e-01   1.61807900e+03]

How would it be possible to make this to work, by printing all the iterations ? Thanks

You have a list of arrays... and then a list of temperatures. — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Dec 5, 2016 at 21:53
probably parameters[i] = parameters[i].append(temperatures[i]). this should append ith element of temperatures to ith element in parameters — Artur Rychlewicz
– Artur Rychlewicz, Commented Dec 5, 2016 at 21:59

hpaulj · Accepted Answer · 2016-12-05 22:21:17Z

3

You have a list of arrays, plus another list or array:

In [656]: parameters = [np.array([1,2,3]) for _ in range(5)]
In [657]: temps=np.arange(5)

to combine them just iterate through (a list comprehension works fine for that), and perform a concatenate (array append) for each pair.

In [659]: [np.concatenate((arr,[t])) for arr, t in zip(parameters, temps)]
Out[659]: 
[array([1, 2, 3, 0]),
 array([1, 2, 3, 1]),
 array([1, 2, 3, 2]),
 array([1, 2, 3, 3]),
 array([1, 2, 3, 4])]

the use append saves us two pairs of [], otherwise it is the same:

 [np.append(arr,t) for arr, t in zip(parameters,temps)]

A clean 'in-place' version:

for i,(arr,t) in enumerate(zip(parameters,temps)):
    parameters[i] = np.append(arr,t)

================

If the subarrays are all the same length, you could turn parameters into a 2d array, and concatenate the temps:

In [663]: np.hstack((np.vstack(parameters),temps[:,None]))
Out[663]: 
array([[1, 2, 3, 0],
       [1, 2, 3, 1],
       [1, 2, 3, 2],
       [1, 2, 3, 3],
       [1, 2, 3, 4]])

edited Dec 5, 2016 at 22:21

answered Dec 5, 2016 at 21:59

hpaulj

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5 Comments

juanpa.arrivillaga Over a year ago

Well, you are creating new arrays anyway. It would serve to point out that this is a poor use of numpy data structures.

juanpa.arrivillaga Over a year ago

The inefficiency in time isn't really due to the list comprehension but the fact that np.array s are being used where probably plain Python list is the way to go.

hpaulj Over a year ago

If you are starting with a list, it is often faster to stick with list operations. Creating an array has a substantial overhead. We have to perform timings on realistic data to tell which is faster.

juanpa.arrivillaga Over a year ago

Right, I was responding to the other person who deleted their comment. "appending" to an array will almost certainly be slower, thought

DavidC. Over a year ago

@hpaulj Thank you very much for your three suggestions: 1) concatenate, 2) np.append (both in one line and in a for loop) and 3) hstack. Sorry for the 20h delay on the answer, but I have been digging hard into all these solutions. I have tried to understand all the syntax and how all these functions behave. First starting with enumerate, followed by zip, and then how to combine these two. I have tried to extrapolate the [np.append(arr,t) for arr, t in zip(parameters,temperatures)] solution to the clean 'in-place' solution: Please see the EDIT

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