List of lists counting values-CodePudding

my_list = [
    ['user1', True, '2022-05-31 05:35:48'],
    ['user2', False, '2022-04-26 00:01:12'],
    ['user3', True, '2022-03-09 14:14:12'],
    ['user3', False, '2022-02-28 09:19:48'],
    ['user5', False, '2022-02-07 18:41:48']
]

I'm trying to figure out how to count the number of times true/false are in this list of lists. Ideally I'd like to receive a structure like the following, thank you.

username counted_trues counted_falses

final_list = [
  ['user1', 1, 0],
  ['user2', 0, 1],
  ['user3', 1, 1],
  ['user5', 0, 1]
  ]

CodePudding user response：

you can use defaultdict for easy implementation

from collections import defaultdict

my_list = [
    ['user1', True, '2022-05-31 05:35:48'],
    ['user2', False, '2022-04-26 00:01:12'],
    ['user3', True, '2022-03-09 14:14:12'],
    ['user3', False, '2022-02-28 09:19:48'],
    ['user5', False, '2022-02-07 18:41:48']
]

class answer_counter:
    def __init__(self):
        self.pos = 0
        self.neg = 0

user2counter = defaultdict(answer_counter)

for entry in my_list:
    user, ans, _ = entry
    if ans:
        user2counter[user].pos  = 1
    else:
        user2counter[user].neg  = 1

final_list = []
for user in user2counter:
    final_list.append([user, user2counter[user].pos, user2counter[user].neg])

"""result
[
['user1', 1, 0], 
['user2', 0, 1], 
['user3', 1, 1], 
['user5', 0, 1]
]

"""

CodePudding user response：

import pandas as pd

pd.DataFrame(my_list).groupby([0, 1]).size().unstack(fill_value=0).T.to_dict()

why don't use pandas?

output:

{'user1': {False: 0, True: 1},
 'user2': {False: 1, True: 0},
 'user3': {False: 1, True: 1},
 'user5': {False: 1, True: 0}}

CodePudding user response：

This code should work:

my_list = [
    ['user1', True, '2022-05-31 05:35:48'],
    ['user2', False, '2022-04-26 00:01:12'],
    ['user3', True, '2022-03-09 14:14:12'],
    ['user3', False, '2022-02-28 09:19:48'],
    ['user5', False, '2022-02-07 18:41:48']
]


output = {}
for user_info in my_list:
    if user_info[0] not in output:
        output[user_info[0]] = [0, 0]
    output[user_info[0]][0]  = user_info[1]
    output[user_info[0]][1]  = not(user_info[1])

final_list = []
for key, value in output.items():
    final_list.append([key, *value])

print(final_list)

It converts the original list (my_list) into a dictionary with the user's name as the key and a list of two values, the number of Trues and the number of Falses. This dictionary is then converted into a list.

CodePudding user response：

You can use comprehension lists:

final_list = [
    [l[0], int(l[1]), int(not l[1])]
    for l in my_list
]