list with string values to 2D array

Question

A list of string values looks like this:

x = ["0: ['17' '19']", "1: ['32' '35']", "2: ['29']", "3: ['16']", "4: ['24' '18' '9']", "6: ['24' '26']", "9: ['11' '26' '34']", "10: ['33']"]

I want a 2D array so I can do this:

print(x[0][1][1])
19

First I get of rid of the colon:

x = [i.split(': ') for i in x]
[['0', "['17' '19']"], ['1', "['32' '35']"], ['2', "['29']"], ['3', "['16']"], ['4', "['24' '18' '9']"], ['6', "['24' '26']"], ['9', "['11' '26' '34']"], ['10', "['33']"]]

But I don't know what to do next ...

Rakesh · Accepted Answer · 2020-05-11 09:51:17Z

1

This is one approach.

Ex:

x = ["0: ['17' '19']", "1: ['32' '35']", "2: ['29']", "3: ['16']", "4: ['24' '18' '9']", "6: ['24' '26']", "9: ['11' '26' '34']", "10: ['33']"]
res = []
for i in x:
    m, n = i.split(": ")
    res.append([m, [int(j.strip("'")) for j in n.strip("[]").split()]])

print(res[0][1][1]) #-->19

Or using numpy

import numpy as np

res = []
for i in x:
    m, n = i.split(": ")
    res.append([m, np.fromstring(n[1:-1].replace("'", ""),sep=' ').astype(int)])

print(res[0][1][1])

edited May 11, 2020 at 9:51

answered May 11, 2020 at 9:46

Rakesh

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Mustafa Aydın · Accepted Answer · 2020-05-11 10:16:02Z

0

import re
the_list = ["0: ['17' '19']", "1: ['32' '35']", "2: ['29']"]

new_list = []
for entry in the_list:
    idx, *vals = map(int, re.findall(r"\d+", entry))
    new_list.append([idx, vals])

print(new_list, new_list[0][1][1], sep="\n")
# [[0, [17, 19]], [1, [32, 35]], [2, [29]]]
# 19

The simple regex \d+ extracts all the numbers in an entry of the list you're looking for as a list e.g. ['1', '32', '35']. Then we map these to integers and unpack it to the index and the remaining values e.g idx = 1 and vals = [32, 35]. Then store for further use.

edited May 11, 2020 at 10:16

answered May 11, 2020 at 10:04

Mustafa Aydın

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