Python 如何在不使用for循环命令的情况下移动数据帧行

Python 如何在不使用for循环命令的情况下移动数据帧行,python,dataframe,for-loop,Python,Dataframe,For Loop,我有以下数据帧: df = pd.DataFrame({'Day' : ['15', '15', '15', '16', '16', '17', '17', '17', '17'], 'Month' : ['10', '10', '10', '10', '10', '10', '10', '10', '10'], 'Year' : ['2019', '2019', '2019

我有以下数据帧:

       df = pd.DataFrame({'Day' : ['15', '15', '15', '16', '16', '17', '17', '17', '17'], 
                          'Month' : ['10', '10', '10', '10', '10', '10', '10', '10', '10'], 
                          'Year' : ['2019', '2019', '2019', '2019', '2019', '2019', '2019', '2019', 
                                    '2019'], 
                          'Hour' : ['14', '14', '14', '14', '14', '14', '15', '15', '15'],
                          'Minute' : ['33', '41', '45', '46', '58', '59', '01', '02', '03' ],
                          'Second' : ['16', '17', '19', '19', '20', '0', '0', '0', '0'],
                          'depth' : [40000, 39000, 13000, 40000, 39500, 35000, 34500, 35000, 34600]
                          })
我使用此行创建了一个新的日期列:

        df['Date'] = pd.to_datetime(df[['Year', 'Month', 'Day', 'Hour', 'Minute', 'Second']])
时间差和深度差受到限制。因此,我实现了以下代码:

      df['Status'] = np.NaN

      for i in range(0, len(df)):    
          for j in range(i+1, len(df)):       

              date_init = pd.to_datetime(df['Date'].iloc[i])
              date_next = pd.to_datetime(df['Date'].iloc[j])


              if(abs(date_init -  date_next) < pd.to_timedelta('0 days 00:10:00')): # 10 minutes          

                 #Calculate the depth variation
                 var_delta_sensor = abs(df['depth'].iloc[i] - df['depth'].iloc[j])


                 if(var_delta_sensor < 1500):

                      #The depth is valid let's accept
                      df['Status'].iloc[i] = 'ACCEPT'
                      df['Status'].iloc[j] = 'ACCEPT'

                  else:
                      #Entering here means that the depth is not valid
                      print("NOT depth")             

              else:
                  #The difference between element i and j is greater than 10 minutes
                  i = j
                  j = i + 1 

在代码中没有重大更改的情况下,如果只将a l'une与下一个进行比较,则可以更改嵌套循环:

对于范围内的i(0,len(df)):
对于范围(i+1,len(df))内的j:

作者:

这将使你节省大量的发生

另一种解决方案是使用diff函数计算var_时间和var_传感器。

例如:

将熊猫作为pd导入
将numpy作为np导入
导入日期时间
df['var_time']=df['Date'].diff()
df['var_delta_sensor']=df['depth'].diff().abs()
时间增量=日期时间。时间增量(分钟=10)#10分钟
df['status']=np。其中(((df['var_time']       print(df)

      Day   Month   Year    Hour    Minute  Second  depth   Date                Status
      15    10      2019    14        33    16      40000   2019-10-15 14:33:16 ACCEPT
      15    10      2019    14        41    17      39000   2019-10-15 14:41:17 ACCEPT
      15    10      2019    14        45    19      13000   2019-10-15 14:45:19 NaN
      16    10      2019    14        46    19      40000   2019-10-16 14:46:19 NaN
      16    10      2019    14        58    20      39500   2019-10-16 14:58:20 NaN
      17    10      2019    14        59    0       35000   2019-10-17 14:59:00 ACCEPT
      17    10      2019    15        01    0       34500   2019-10-17 15:01:00 ACCEPT
      17    10      2019    15        02    0       35000   2019-10-17 15:02:00 ACCEPT
      17    10      2019    15        03    0       34600   2019-10-17 15:03:00 ACCEPT
 for i in range(0, len(df)-1):    
      j=i+1
import pandas as pd
import numpy as np
import datetime

df['var_time'] =df['Date'].diff()
df['var_delta_sensor'] =df['depth'].diff().abs()

time_delta=datetime.timedelta(minutes=10) #10 minutes          

df['status']  = np.where((( df['var_time'] < time_delta) & (df['var_delta_sensor'] <1500)), 'Accept', np.NaN)