从列表的第一个元素中删除双引号

Question

I want to remove double quotes from this list:

>>> a
["6/1, 0, 0, ['79', '80', '81', '82']"]

I tried converting first element of list to list which is a string but dint work:

>>> a[0]
"6/1, 0, 0, ['79', '80', '81', '82']"

>>> list(a[0])
['6', '/', '1', ',', ' ', '0', ',', ' ', '0', ',', ' ', '[', "'", '7', '9', "'", ',', ' ', "'", '8', '0', "'", ',', ' ', "'", '8', '1', "'", ',', ' ', "'", '8', '2', "'", ']']

Expected output:

>>> a
[6/1, 0, 0, ['79', '80', '81', '82']]
                    or
>>> a
['6/1', '0', '0', ['79', '80', '81', '82']]

Can someone please help me resolving this issue?

Answer 1

This will evaluate the string as a list:

a = list(eval(a[0]))

[6, 0, 0, ['79', '80', '81', '82']]

If you know your port will be 6/1, you can replace it with a string like this:

a = list(eval(a[0]))
a[0] = '6/1'

['6/1', 0, 0, ['79', '80', '81', '82']]

If your port is going to change, but you know it will always be the first element, you can do this:

a = list(eval("'" + a[0].split(',')[0] + "'" + "," + ','.join(a[0].split(',')[1:])))

['6/1', 0, 0, ['79', '80', '81', '82']]

Or cleaner version:

l = a[0].split(',')
a = list(eval("'" + l[0] + "' ," + ','.join(l[1:])))

['6/1', 0, 0, ['79', '80', '81', '82']]

Answer 2

x= ["6/1, 0, 0, ['79', '80', '81', '82']" ]
a=x[0].replace('\'','').split(', [')
finallist=a[0].split(',')
finallist.append(a[1].replace(']','').split(','))
print (finallist)

output

['6/1', ' 0', ' 0', ['79', ' 80', ' 81', ' 82']]