Question

我正在提高脚本的速度，并看到以下答案：Iterrows Performance Issues。在这里，答案说很少需要使用iterrows。

在我的代码中，我使用了iterrows，因为它使用非常简单直观，但也非常慢。所以我想要渲染我使用iterrows的代码片段。这里有两个例子，我无法找到解决方案。在这两个示例中，列中的值都是具有以下格式的日期时间值：%Y-%m-%d %H:%M:%S

for index, row in df.iterrows():
    df.loc[index, 'Time_Between']= row['Time_Begin'] + timedelta(seconds=row['Some_Integer_Seconds_In_A_Column'])
    df.loc[index, 'Time_Required']= row['Time_End'] - timedelta(seconds=SomeIntegerSecondsAsAVariable)
    df.loc[index, 'Tota_Time']= ((row['Time_Begin'] - row['Time_First']).total_seconds())/60


for index, row in df.iterrows():
    if row['Time_Required'] > row['Time_Between']:
        df.loc[index, 'Check']= 0
    else:
        df.loc[index, 'Check']= 1

我该如何对此进行矢量化？我试着掩盖和申请，但我无法得到任何工作。我得到的大部分时间都是：TypeError: Cannot change data-type for object array.我不知道的东西......

Answer 1

我认为你可以使用：

import pandas as pd
import numpy as np

df = pd.DataFrame({'Time_End': {0: pd.Timestamp('2015-11-15 00:00:00'), 1: pd.Timestamp('2015-10-18 00:00:00'), 2: pd.Timestamp('2015-10-17 00:00:00'), 3: pd.Timestamp('2015-10-16 00:00:00')}, 'Int_Sec': {0: 4, 1: 2, 2: 7, 3: 10}, 'Time_First': {0: pd.Timestamp('2015-10-15 00:00:00'), 1: pd.Timestamp('2015-10-15 00:00:00'), 2: pd.Timestamp('2015-12-15 00:00:00'), 3: pd.Timestamp('2015-12-15 00:00:00')}, 'Time_Begin': {0: pd.Timestamp('2015-10-15 10:00:00'), 1: pd.Timestamp('2015-10-15 12:00:00'), 2: pd.Timestamp('2015-12-15 10:00:00'), 3: pd.Timestamp('2015-12-15 10:00:00')}})
print (df)
   Int_Sec          Time_Begin   Time_End Time_First
0        4 2015-10-15 10:00:00 2015-11-15 2015-10-15
1        2 2015-10-15 12:00:00 2015-10-18 2015-10-15
2        7 2015-12-15 10:00:00 2015-10-17 2015-12-15
3       10 2015-12-15 10:00:00 2015-10-16 2015-12-15

Sec_Var = 20
df['Time_Between'] = df['Time_Begin'] + pd.to_timedelta(df['Int_Sec'], unit='s')
df['Time_Required'] = df['Time_End'] - pd.to_timedelta(Sec_Var, unit='s')
df['Tota_Time'] = ((df['Time_Begin'] - df['Time_First']).dt.total_seconds()) / 60

df['Check'] = np.where(df['Time_Required'] > df['Time_Between'], 0, 1)

print (df)

   Int_Sec          Time_Begin   Time_End Time_First        Time_Between  \
0        4 2015-10-15 10:00:00 2015-11-15 2015-10-15 2015-10-15 10:00:04   
1        2 2015-10-15 12:00:00 2015-10-18 2015-10-15 2015-10-15 12:00:02   
2        7 2015-12-15 10:00:00 2015-10-17 2015-12-15 2015-12-15 10:00:07   
3       10 2015-12-15 10:00:00 2015-10-16 2015-12-15 2015-12-15 10:00:10   

        Time_Required  Tota_Time  Check  
0 2015-11-14 23:59:40      600.0      0  
1 2015-10-17 23:59:40      720.0      0  
2 2015-10-16 23:59:40      600.0      1  
3 2015-10-15 23:59:40      600.0      1

Answer 2

第二集：

public async static Task<string> Something()
    {            
        var http = new HttpClient();
        var url = "https://maps.googleapis.com/maps/api/geocode/json?address=Los%20Angeles,CA=AIzaSyDABt";
        var response = await http.GetAsync(url);
        if (response.IsSuccessStatusCode)
        {
            var result = await response.Content.ReadAsStringAsync();
            result = JsonConvert.DeserializeObject<string>(result);
            return result;
        }
        return "";
    }

var result = Task.Run(() => Something()).Result;

第一集：

for index, row in df.iterrows():
    if row['Time_Required'] > row['Time_Between']:
        df.loc[index, 'Check']= 0
    else:
       df.loc[index, 'Check']= 1

np.where(row['Time_Required'] > row['Time_Between'], df['Check']= 0, df['Check']= 1)

矢量化替代iterrows

2 个答案: