How to access the first items in a list of arrays without for loop in Python 3?

Question

I have an ode equation which produces a list like below:

u=[array([ 2.06642033, -0.03448756]), 
   array([ 2.03964994, -0.18737285]), 
   array([ 1.99884859, -0.21461016]), 
   array([ 1.95476809, -0.2254584 ]), 
   array([ 1.90875336, -0.23472173]), 
   array([ 1.86082857, -0.24471069]),
   ... ]

I want to plot u[0] and u[1] based on time. I try to access all first and second elements of u in two list with l1=u[0:len(u)-1][0] and l2=u[0:len(u)-1][1], but it gives me only the first item from list. Does anyone have a solution for it? Thanks

What do you mean when you say "based on time"? Please provide some examples. — Alessandro Romano
– Alessandro Romano, Commented Apr 9, 2018 at 9:39
I have a time list which has the same length as u, like this: t=[0,0.001, ...] — Maryam
– Maryam, Commented Apr 9, 2018 at 9:40

jpp · Accepted Answer · 2018-04-09 09:40:41Z

4

You should convert your list of arrays into a single numpy array.

Also note that numpy array indexing in 2d is performed by arr[row, column]. If you don't filter by a dimension, just use :.

from numpy import array

u = [array([ 2.06642033, -0.03448756]), array([ 2.03964994, -0.18737285]),
     array([ 1.99884859, -0.21461016]), array([ 1.95476809, -0.2254584 ]),
     array([ 1.90875336, -0.23472173]), array([ 1.86082857, -0.24471069])]

u = np.array(u)

res = u[:, 0]

# array([ 2.06642033,  2.03964994,  1.99884859,  1.95476809,  1.90875336,
#         1.86082857])

answered Apr 9, 2018 at 9:40

jpp

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