问题描述
我有这样的事情：
fromJobtitle toJobtitle size
0 CEO CEO 65
1 CEO Vice President 23
2 CEO Employee 56
3 Vice President CEO 112
4 Employee CEO 20
我想计算共现的次数，以便它结合两次出现（仅显示 2 之间有多少元素）
输出示例：
0 CEO Vice President 135
1 CEO Employee 76
2 CEO CEO 65
解决方法
Error: Output stream closed
at Timeout._onTimeout (C:\Users\salos\OneDrive\Desktop\Speech-Recognition-Bot\node_modules\fluent-ffmpeg\lib\processor.js:491:25)
at listOnTimeout (node:internal/timers:557:17)
at processTimers (node:internal/timers:500:7)
Emitted 'error' event on FfmpegCommand instance at:
at emitEnd (C:\Users\salos\OneDrive\Desktop\Speech-Recognition-Bot\node_modules\fluent-ffmpeg\lib\processor.js:424:16)
at Timeout._onTimeout (C:\Users\salos\OneDrive\Desktop\Speech-Recognition-Bot\node_modules\fluent-ffmpeg\lib\processor.js:491:17)
at listOnTimeout (node:internal/timers:557:17)
at processTimers (node:internal/timers:500:7)
import pandas as pd
df = pd.DataFrame({
'fromJobtitle': ['CEO','CEO','Vice President','Employee'],'toJobtitle': ['CEO','Employee','CEO'],'size': [65,23,56,112,20]
})
然后：
df['combination'] = df.apply(lambda row: tuple(sorted([
row['fromJobtitle'],row['toJobtitle']
])),axis=1)
结果：
df = df.groupby('combination').sum().reset_index()
最后：
combination size
0 (CEO,CEO) 65
1 (CEO,Employee) 76
2 (CEO,Vice President) 135
结果：
df['from'] = df.apply(lambda row: row['combination'][0],axis=1)
df['to'] = df.apply(lambda row: row['combination'][1],axis=1)
df = df.drop('combination',axis=1)
df.head()
试试：
func application(_ application: UIApplication,didRegisterForRemoteNotificationsWithDeviceToken deviceToken: Data) {
Messaging.messaging().apnsToken = deviceToken
}
结果如下：
df.groupby(lambda x: tuple(sorted(df.loc[x,['fromJobTitle','toJobTitle']]))).sum()
这是一个不同的解决方案：
首先创建一个按字母顺序组合名称的列
"a" -> true
"aa" -> true
"abb" -> true
"aaa" -> true
"abba" -> true
然后按该名称分组并求和
df['titles'] = np.where(df['fromJobtitle']<df['toJobtitle'],df['fromJobtitle']+"|"+df['toJobtitle'],df['toJobtitle']+"|"+df['fromJobtitle'])
0 CEO|CEO
1 CEO|Vice President
2 CEO|Employee
3 CEO|Vice President
4 CEO|Employee
Name: titles,dtype: object
然后只需将组合名称拆分为单独的部分
df_groups = df.groupby('titles').sum().reset_index()
df_groups
titles size
CEO|CEO 65
CEO|Employee 76
CEO|Vice President 135