Question

问候所有人，我的python文件中有一个列表，根据此列表，我如何计算每个员工A，B和C的actual_duration的总和？

以下是我的清单：

[{'employee_name': 'A', 'actual_duration': 14}, 
{'employee_name': 'A', 'actual_duration': 6}, 
{'employee_name': 'B', 'actual_duration': 8}, 
{'employee_name': 'B', 'actual_duration': 22}, 
{'employee_name': 'C', 'actual_duration': 15}]

Answer 1

一种方法是使用collections.defaultdict

l = [   {'employee_name': 'A', 'actual_duration': 14}, 
        {'employee_name': 'A', 'actual_duration': 6}, 
        {'employee_name': 'B', 'actual_duration': 8}, 
        {'employee_name': 'B', 'actual_duration': 22}, 
        {'employee_name': 'C', 'actual_duration': 15}  ]

from collections import defaultdict
d_dict = defaultdict(int)

for k,v in [d.values() for d in l]:
    d_dict[k]+=v

new_l = []
for k,v in d_dict.items():
    new_l.append({'employee_name': k, 'actual_duration': v})

new_l

输出：

[{'actual_duration': 20, 'employee_name': 'A'},
 {'actual_duration': 30, 'employee_name': 'B'},
 {'actual_duration': 15, 'employee_name': 'C'}]

Answer 2

尝试以下操作，使用for循环迭代原始列表并连接项目：

for entry in times:
    if entry['employee_name'] not in map(lambda i: i['employee_name'], aggregate):
            aggregate.append(entry)
    else:
            aggregate[map(lambda i: i['employee_name'], aggregate).index(entry['employee_name'])]['actual_duration']+=entry['actual_duration']

>>> times = [{'employee_name': 'A', 'actual_duration': 14}, 
... {'employee_name': 'A', 'actual_duration': 6}, 
... {'employee_name': 'B', 'actual_duration': 8}, 
... {'employee_name': 'B', 'actual_duration': 22}, 
... {'employee_name': 'C', 'actual_duration': 15}]
>>> aggregate = []
>>> for entry in times:
...     if entry['employee_name'] not in map(lambda i: i['employee_name'], aggregate):
...             aggregate.append(entry)
...     else:
...             print map(lambda i: i['employee_name'], aggregate).index(entry['employee_name'])
...             aggregate[map(lambda i: i['employee_name'], aggregate).index(entry['employee_name'])]['actual_duration']+=entry['actual_duration']
... 
0
1
>>> aggregate
[{'actual_duration': 20, 'employee_name': 'A'}, {'actual_duration': 30, 'employee_name': 'B'}, {'actual_duration': 15, 'employee_name': 'C'}]
>>>

Answer 3

有两种方法，一种是使用默认字典：

your_data=[{'employee_name': 'A', 'actual_duration': 14},
{'employee_name': 'A', 'actual_duration': 6},
{'employee_name': 'B', 'actual_duration': 8},
{'employee_name': 'B', 'actual_duration': 22},
{'employee_name': 'C', 'actual_duration': 15}]


from collections import defaultdict

default_dict=defaultdict(list)

for i in your_data:
    track=[]
    for j,l in i.items():
        [track.insert(0, (l, j)) if isinstance(l,int) else track.insert(1,(l,j))]





    default_dict[track[1]].append(track[0][0])

print({'employee_name : {}'.format(key[0]):'actual_duration : {} '.format(sum(values)) for key,values in default_dict.items()})

输出：

{'employee_name : A': 'actual_duration : 20 ', 'employee_name : C': 'actual_duration : 15 ', 'employee_name : B': 'actual_duration : 30 '}

不使用default_dict的第二个详细解决方案：

new={}
for i in your_data:
    track=[]
    for j,l in i.items():
        [track.insert(0, (l, j)) if isinstance(l,int) else track.insert(1,(l,j))]


    if track[1] not in new:
        new[track[1]]=[track[0][0]]
    else:
        new[track[1]].append(track[0][0])








print({'employee_name : {}'.format(key[0]):'actual_duration : {} '.format(sum(values)) for key,values in new.items()})

Answer 4

只需使用pandas

import pandas as pd

data = [{'employee_name': 'A', 'actual_duration': 14}, 
{'employee_name': 'A', 'actual_duration': 6}, 
{'employee_name': 'B', 'actual_duration': 8}, 
{'employee_name': 'B', 'actual_duration': 22}, 
{'employee_name': 'C', 'actual_duration': 15}]

result = pd.DataFrame(data).groupby('employee_name').sum()
print(result)

               actual_duration
employee_name                 
A                           20
B                           30
C                           15

如果你想要一个字典，你可以做

result_dict = result.reset_index().to_dict('records')
print(result_dict)

[{'actual_duration': 20, 'employee_name': 'A'},
 {'actual_duration': 30, 'employee_name': 'B'},
 {'actual_duration': 15, 'employee_name': 'C'}]

根据员工姓名字段计算python列表中的值之和

4 个答案: