Extract individual values from numpy array into a list

Question

I am trying to extract the values from a numpy array and place the individual entries into a list.

For example, if I have the following list:

import numpy as np
x = 1
problem_variable = np.array(['a', 'b'], dtype='<U9')
z = 2
mylist = [x, problem_variable , z]
# [1, array(['a', 'b'], dtype='<U9'), 2]

How do I get the result

[1, 'a', 'b', 2]

I do not know the length of problem_variable before hand so cannot hard code problem_variable[0] , problem_variable[1] etc.

The following does what I want but I am sure that I am missing the appropriate way to break the array apart. Thanks.

result = []
result.append(x)
for i in problem_variable: result.append(i)
result.append(z)

Answer 1

你可以解压你的数组：

mylist = [x, *problem_variable , z]

Answer 2

Since you're using numpy : you can use np.r_ to concatenate the input objects along first axis:

np.r_[x, problem_variable, z]
# array(['1', 'a', 'b', '2'], dtype='<U9')

---

Comparing performance on larger lists:

problem_var = np.concatenate([problem_variable]*10000, axis=0)

%timeit np.r_[x, problem_var, z]
# 143 µs ± 19.6 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

%timeit np.hstack((x, problem_var , z))
# 553 µs ± 55.5 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit [x, *problem_var.tolist() , z]
# 502 µs ± 42.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit [x, *problem_var , z]
# 6.46 ms ± 215 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Answer 3

You can use np.hstack , which can stack arrays in sequence horizontally,

np.hstack((x, problem_variable , z))

# ['1' 'a' 'b' '2']