Question

我有一个清单，列出了一天中不同时间工作的员工，我想计算每个人工作的天数，例如：

脚：3 BAZ：3 NOM：1个等等。

这是我接收原始数据的方式：

my_list = [('NOM', datetime.date(2030, 1, 1)),
 ('BAR', datetime.date(2019, 4, 8)),
 ('HAM', datetime.date(2019, 4, 8)),
 ('FOO', datetime.date(2019, 4, 8)),
 ('BAZ', datetime.date(2019, 4, 8)),
 ('BAR', datetime.date(2019, 4, 10)),
 ('BAZ', datetime.date(2019, 4, 10)),
 ('FOO', datetime.date(2019, 4, 10)),
 ('HAM', datetime.date(2019, 4, 10)),
 ('HAM', datetime.date(2019, 4, 10)),
 ('FOO', datetime.date(2019, 4, 10)),
 ('BAR', datetime.date(2019, 4, 10)),
 ('BAZ', datetime.date(2019, 4, 10)),
 ('BAZ', datetime.date(2019, 4, 11)),
 ('HAM', datetime.date(2019, 4, 11)),
 ('BAR', datetime.date(2019, 4, 11)),
 ('FOO', datetime.date(2019, 4, 11)),
 ('BAZ', datetime.date(2019, 4, 11)),
 ('FOO', datetime.date(2019, 4, 11)),
 ('BAR', datetime.date(2019, 4, 11)),
 ('HAM', datetime.date(2019, 4, 11)),
 ('FOO', datetime.date(2019, 4, 11)),
 ('BAZ', datetime.date(2019, 4, 11)),
 ('BAR', datetime.date(2019, 4, 11)),
 ('HAM', datetime.date(2019, 4, 11)),
 ('FOO', datetime.date(2019, 4, 11)),
 ('BAZ', datetime.date(2019, 4, 11)),
 ('BAR', datetime.date(2019, 4, 11)),
 ('HAM', datetime.date(2019, 4, 11))]

我设法将每个人的清单剥离为唯一的日期，如下所示： a = Counter(set(dictio))每天消除一个人的重复：

Counter({('HAM', datetime.date(2019, 4, 8)): 1,
         ('HAM', datetime.date(2019, 4, 10)): 1,
         ('HAM', datetime.date(2019, 4, 11)): 1,
         ('BAR', datetime.date(2019, 4, 8)): 1,
         ('BAR', datetime.date(2019, 4, 10)): 1,
         ('BAR', datetime.date(2019, 4, 11)): 1,
         ('FOO', datetime.date(2019, 4, 8)): 1,
         ('FOO', datetime.date(2019, 4, 10)): 1,
         ('FOO', datetime.date(2019, 4, 11)): 1,
         ('BAZ', datetime.date(2019, 4, 8)): 1,
         ('BAZ', datetime.date(2019, 4, 10)): 1,
         ('BAZ', datetime.date(2019, 4, 11)): 1,
         ('NOM', datetime.date(2030, 1, 1)): 1})

这就是我遇到的问题：我要从上面转到：

HAM:3
BAR:3
FOO:3
BAZ:3
NOM:1

Answer 1

您可以使用

import collections
collections.Counter(x for x , y in set(my_list) )
Out[251]: Counter({'BAR': 3, 'BAZ': 3, 'FOO': 3, 'HAM': 3, 'NOM': 1})

Answer 2

使用itertools.groupby：

from itertools import groupby
from operator import itemgetter

result = {key: len(group) for key, group in groupby(sorted(set(my_list)), key=itemgetter(0))}
print(result)

这会按第一个元素（名称）对my_list进行排序，根据这些名称将其划分为组，最后得到每个组的名称和长度作为{{1}中的键值对}。

输出：

dict

Answer 3

将列表转换为熊猫数据框，删除重复项并按名称分组

import datetime
import pandas as pd
my_list = [('NOM', datetime.date(2030, 1, 1)),
 ('BAR', datetime.date(2019, 4, 8)),
 ('HAM', datetime.date(2019, 4, 8)),
 ('FOO', datetime.date(2019, 4, 8)),
 ('BAZ', datetime.date(2019, 4, 8)),
 ('BAR', datetime.date(2019, 4, 10)),
 ('BAZ', datetime.date(2019, 4, 10)),
 ('FOO', datetime.date(2019, 4, 10)),
 ('HAM', datetime.date(2019, 4, 10)),
 ('HAM', datetime.date(2019, 4, 10)),
 ('FOO', datetime.date(2019, 4, 10)),
 ('BAR', datetime.date(2019, 4, 10)),
 ('BAZ', datetime.date(2019, 4, 10)),
 ('BAZ', datetime.date(2019, 4, 11)),
 ('HAM', datetime.date(2019, 4, 11)),
 ('BAR', datetime.date(2019, 4, 11)),
 ('FOO', datetime.date(2019, 4, 11)),
 ('BAZ', datetime.date(2019, 4, 11)),
 ('FOO', datetime.date(2019, 4, 11)),
 ('BAR', datetime.date(2019, 4, 11)),
 ('HAM', datetime.date(2019, 4, 11)),
 ('FOO', datetime.date(2019, 4, 11)),
 ('BAZ', datetime.date(2019, 4, 11)),
 ('BAR', datetime.date(2019, 4, 11)),
 ('HAM', datetime.date(2019, 4, 11)),
 ('FOO', datetime.date(2019, 4, 11)),
 ('BAZ', datetime.date(2019, 4, 11)),
 ('BAR', datetime.date(2019, 4, 11)),
 ('HAM', datetime.date(2019, 4, 11))]

# COnvert List of Tuples to Dataframe
df = pd.DataFrame(my_list,columns=['name','date'])
#Remove Duplicates
df.drop_duplicates(inplace=True)
#Group by Name Count
df.groupby('name').count()

计算工作天数，即计算python列表中的项目

3 个答案: