I have a string extracted from a .csv which has this format:
str = "[point, contextual, point]"
What I wanna do is convert it to a list in the format:
str = ["point", "contextual", "point"]
How can I do it? I tried with json.loads(str) but I got the error:
json.decoder.JSONDecodeError: Expecting value: line 1 column 2 (char 1)
you could use:
my_str = "[point, contextual, point]"
my_str[1:-1].split(', ') # remove first and last characters ([]) and split on ", "
NB. don't use str as a variable name, this overwrites the str
builtin
re.split and ,\s* as separatorYou could use this expression : str[1:-1].split(", ")
By the way it is not recommended to give a variable the name of a python type.
I suggest you use yaml:
import yaml
s = "[point, contextual, point]"
s = yaml.load(s, Loader=yaml.FullLoader)
print(s)
Output
['point', 'contextual', 'point']
Note:
yaml is a third party module to install it do (at the command line):
pip install pyyaml
The yaml version of the code snippet above is 5.4.1
Because that's not a valid JSON string, the json.loads doesn't work. You need to do it manually and you can not use json module for the string you have.
st = "[point, contextual, point]"
lst = st[1:-1].split(', ')
On a side note, don't use str as a vairable: that's a builtin
Try this code:
string = "[point, contextual, point]"
print(string[1:-1].split(', '))
Outputs:
['point', 'contextual', 'point']
Tell me if its okay for you...
split already returns a listYou can try this:
>>> st = "[point, contextual, point]"
>>> st[1:-1]
[point, contextual, point]
>>> st[1:-1].split(',')
['point', ' contextual', ' point'] # <- you have space
>>> list(map(lambda x: x.strip(), st[1:-1].split(',')))
['point', 'contextual', 'point']
why not split(', ')?
>>> st = "[point, contextual, point]"
>>> st[1:-1].split(', ')
['point', ' contextual', ' point']
>>> list(map(lambda x: x.strip(), st[1:-1].split(',')))
['point', 'contextual', 'point']
', ' instead?
stras that's the name of the built-in string type.