Convert string type array to array

Question

I have this:

[s[8] = 5,

 s[4] = 3,

 s[19] = 2,

 s[17] = 8,

 s[16] = 8,

 s[2] = 8,

 s[9] = 7,

 s[1] = 2,

 s[3] = 9,

 s[15] = 7,

 s[11] = 0,

 s[10] = 9,

 s[12] = 3,

 s[18] = 1,

 s[0] = 4,

 s[14] = 5,

 s[7] = 4,

 s[6] = 2,

 s[5] = 7,

 s[13] = 9]

How can I turn this into a python array where I can do for items in x: ?

What kind of data-structure is that? It has the list brackets around it but contains notation to set key and value in a dict? — Maurice, Sep 22 '16 at 18:35
here is solved example with [numpy](http://stackoverflow.com/questions/3877209/how-to-convert-an-array-of-strings-to-an-array-of-floats-in-numpy) — Ari Gold, Sep 22 '16 at 18:35
Then edit you answer to mentioned it as string. `"` around each item — Moinuddin Quadri, Sep 22 '16 at 18:36
Its definitely a string, couldn't edit it, it kept showing me error that its not formatted. So I can't add quotes. Expected output is as I said an array, S array where I can `print s[0]` and get it's value. S string should turn into actual python array — , Sep 22 '16 at 18:40
Just to be clear, are you looking for an [array](https://docs.python.org/3.6/library/array.html), or a [list](https://docs.python.org/3.6/library/stdtypes.html#list)? — Kevin, Sep 22 '16 at 18:43
You want us to turn that string into a dictionary with the left side of the equation as the key and the right side as the value? — Maurice, Sep 22 '16 at 18:43
I am assuming, you have the above list, let's say `my_list` and another list as `s`. On doing `my_list[0]`, it should execute `s[8] = 5` and set `8`th index of `s` as `8`. Is it the case? — Moinuddin Quadri, Sep 22 '16 at 18:46
Not relevant to the question, but I'd suggest changing the communication format if possible. — Pavel Gurkov, Sep 22 '16 at 18:48

score 4 · Accepted Answer · answered Sep 22 '16 at 18:47

import re

data = """[s[8] = 5,

 s[4] = 3,

 s[19] = 2,

 s[17] = 8,

 s[16] = 8,

 s[2] = 8,

 s[9] = 7,

 s[1] = 2,

 s[3] = 9,

 s[15] = 7,

 s[11] = 0,

 s[10] = 9,

 s[12] = 3,

 s[18] = 1,

 s[0] = 4,

 s[14] = 5,

 s[7] = 4,

 s[6] = 2,

 s[5] = 7,

 s[13] = 9]"""

d = {int(m.group(1)): int(m.group(2)) for m in re.finditer(r"s\[(\d*)\] = (\d*)", data)}
seq = [d.get(x) for x in range(max(d))]
print(seq)
#result: [4, 2, 8, 9, 3, 7, 2, 4, 5, 7, 9, 0, 3, 9, 5, 7, 8, 8, 1]

score 1 · Answer 2 · answered Sep 22 '16 at 19:08

If you want the array to have the same name that is in the input string, you could use exec. This is not very pythonic, but it works for simple stuff

string = ("[s[8] = 5, s[4] = 3, s[19] = 2,"
   "s[17] = 8, s[16] = 8, s[2] = 8,"
   "s[9] = 7,  s[1] = 2,  s[3] = 9,"
   "s[15] = 7, s[11] = 0, s[10] = 9,"
   "s[12] = 3, s[18] = 1, s[0] = 4,"
   "s[14] = 5, s[7] = 4,  s[6] = 2,"
   "s[5] = 7, s[13] = 9]")

items = [item.rstrip().lstrip() for item in string[1:-1].split(",")]
name = items[0].partition("[")[0]
# Create the array
exec("{} = [None] * {}".format(name, len(items)))
# Populate with the values of the string
for item in items:
    exec(items[0].partition("[")[0] )

This will generate an array named "s" and if there is an index missing it will be initialized as None

score 0 · Answer 3 · answered Sep 22 '16 at 18:51

Assuming this is one giant string, if you want print s[0] to print 4, then you need to split this up by the commas, then iterate through each item.

inputArray = yourInput[1:-1].replace(' ','').split(',\n\n')
endArray = [0]*20
for item in inputArray:
    endArray[int(item[item.index('[')+1:item.index(']')])]= int(item[item.index('=')+1:])
print endArray

score 0 · Answer 4 · answered Sep 22 '16 at 18:55

An easy way, although less efficient than Kevin's solution (without using regular expressions), would be the following (where some_array is your string):

sub_list = some_array.split(',')
some_dict = {}

for item in sub_list:
    sanitized_item = item.strip().rstrip().lstrip().replace('=', ':')

    # split item in key val
    k = sanitized_item.split(':')[0].strip()
    v = sanitized_item.split(':')[1].strip()

    if k.startswith('['):
        k = k.replace('[', '')

    if v.endswith(']'):
        v = v.replace(']', '')

    some_dict.update({k: int(v)})

print(some_dict)
print(some_dict['s[9]'])

Output sample:

{'s[5]': 7, 's[16]': 8, 's[0]': 4, 's[9]': 7, 's[2]': 8, 's[3]': 9, 's[10]': 9, 's[15]': 7, 's[6]': 2, 's[7]': 4, 's[14]': 5, 's[19]': 2, 's[17]': 8, 's[4]': 3, 's[12]': 3, 's[11]': 0, 's[13]': 9, 's[18]': 1, 's[1]': 2, 's8]': 5}
7

Convert string type array to array

4 Answers4