How to convert a list to a dict?

Question

I am using subprocess to print the output of ls .

output = subprocess.getoutput("ssh -i key.pem ubuntu@10.127.6.83 ls -l --time-style=long-iso /opt/databases | awk -F' ' '{print $6 $8}'")
lines = output.splitlines()
print(lines)
format = '%Y-%m-%d'
for line in lines:
   if line != '':
      date = datetime.strptime(line, format)

And when I print lines am getting a large list in the following format:

['', '2019-04-25friendship_graph_43458', '2019-07-18friendship_graph_45359', '2019-09-03friendship_graph_46553', '2019-10-02friendship_graph_46878']

I am trying to convert the above output to a dict with the dates in '%Y-%m-%d' format. So output would be something like:

{ '2019-04-25' : 'friendship_graph_43458',
  '2019-07-18': 'friendship_graph_45359',
  '2019-09-03': 'friendship_graph_46553' }

and so on, but not quite sure how to do so.

Answer 1

Technically if you don't want to use re if all dates are formatted the same then they will all be 10 characters long thus just slice the strings to make the dict in a comprehension:

data = ['', '2019-04-25friendship_graph_43458', '2019-07-18friendship_graph_45359', '2019-09-03friendship_graph_46553', '2019-10-02friendship_graph_46878']

output = {s[:10]: s[10:] for s in data if len(s) > 10}

---

{'2019-04-25': 'friendship_graph_43458', '2019-07-18': 'friendship_graph_45359', '2019-09-03': 'friendship_graph_46553', '2019-10-02': 'friendship_graph_46878'}

Answer 2

You could use a regular expression for each item in the list. For example:

(\d{4}-\d{2}-\d{2})(.*)

Then, you can just iterate through each item in the list and use the regular expression to the get the string in its two parts.

>>> import re
>>> regex = re.compile(r"(\d{4}-\d{2}-\d{2})(.*)")
>>> items = ['', '2019-04-25friendship_graph_43458', '2019-07-18friendship_graph_45359', '2019-09-03friendship_graph_46553', '2019-10-02friendship_graph_46878']
>>> items_dict = {}
>>> for i in items:
        match = regex.search(i)
        if match is None:
            continue
        items_dict[match.group(1)] = match.group(2)

    
>>> items_dict
{'2019-04-25': 'friendship_graph_43458', '2019-07-18': 'friendship_graph_45359', '2019-09-03': 'friendship_graph_46553', '2019-10-02': 'friendship_graph_46878'}

Answer 3

For lines that start with the date; use slices to separate the key from the value.

>>> s = '2019-04-25friendship_graph_43458'
>>> d = {}
>>> d[s[:10]] = s[10:]
>>> d
{'2019-04-25': 'friendship_graph_43458'}
>>>

Answer 4

Use re.findall and dictionary comprehension :

import re
lst = ['', '2019-04-25friendship_graph_43458', '2019-07-18friendship_graph_45359', '2019-09-03friendship_graph_46553', '2019-10-02friendship_graph_46878']
dct = {k: v for s in lst for k, v in re.findall(r'(\d\d\d\d-\d\d-\d\d)(.*)', s) }
print(dct)
# {'2019-04-25': 'friendship_graph_43458', '2019-07-18': 'friendship_graph_45359', '2019-09-03': 'friendship_graph_46553', '2019-10-02': 'friendship_graph_46878'}