初始化列表的dict并在Python中更新其值

Question

I have a data frame below:

user | speed
------------
Anna | 1.0
Bell | 1.2
Anna | 1.3
Chad | 1.5
Bell | 1.4
Anna | 1.1

that I want to use a dictionary to keep record of the number of encounters for each user and update his/her speed when I loop through the data frame.

For instance, the first time we see "Anna" the dictionary is:

{"Anna": [1, 1.0]}

and the second time we see "Anna" it becomes:

{"Anna": [2, 1.3], "Bell": [1, 1.2]}

in the end the dictionary should be:

{"Anna": [3, 1.1], "Bell": [2, 1.4], "Chad": [1, 1.5]}

The counting part is easy:

>>> import pandas as pd
>>> record = pd.DataFrame({"user": ("Anna", "Bell", "Anna", "Chad", "Bell", "Anna"), "speed": (1.0, 1.2, 1.3, 1.5, 1.4, 1.1)})
>>> record
   speed  user
0    1.0  Anna
1    1.2  Bell
2    1.3  Anna
3    1.5  Chad
4    1.4  Bell
5    1.1  Anna
>>> encounter = {}
>>> for i in record['user']:
...   encounter[i] = encounter.get(i, 0) + 1
...
>>> encounter
{'Anna': 3, 'Bell': 2, 'Chad': 1}

but what's a good way to create an empty dictionary of list and update the second value? Thanks!

Answer 1

I believe this is what you want in two lines.

import pandas as pd

record = pd.DataFrame({
  "user": ("Anna", "Bell", "Anna", "Chad", "Bell", "Anna"), 
  "speed": (1.0, 1.2, 1.3, 1.5, 1.4, 1.1)
})

encounter = {}
for name, value in zip(record["user"], record["speed"]):
  encounter[name] = [encounter.get(name, [0])[0] + 1, value]

1. The zip method let you loop through the names and speeds simultaneously.
2. The get method try gets the record if it exists, else return a list [0] .
3. The second [0] takes the first element of the list, which is the counter.
4. Finally it combines with the new value, becomes a new list, and assigned to encounter[name] .

Answer 2

Using collections.Counter

Ex:

import pandas as pd
from decimal import Decimal
from collections import Counter

record = pd.DataFrame({"user": ("Anna", "Bell", "Anna", "Chad", "Bell", "Anna"), "speed": (1.0, 1.2, 1.3, 1.5, 1.4, 1.1)})

encounter = {}
for k,v in Counter(record["user"].tolist()).items():
    encounter[k] = [v, (record[record["user"] == k]["speed"].iloc[-1]).round(1).astype(Decimal)]

print(encounter)

Output:

{'Anna': [3, 1.1], 'Chad': [1, 1.5], 'Bell': [2, 1.4]}

Answer 3

go with pandorable ,

my_dictionary={}
for k, v in df.groupby('user'):
    my_dictionary[k]=[len(v),v.iloc[-1]['speed']]
print(my_dictionary)
{'Anna': [3, 1.1], 'Bell': [2, 1.4], 'Chad': [1, 1.5]}