熊猫的迭代跳转到新的索引，行继续迭代

如何解决熊猫的迭代跳转到新的索引，行继续迭代

大事

我知道进行排序可能不是最好的方法，因此我也对其他想法持开放态度，但我真的很想知道如何执行以下操作：

说我有一个循环（循环！），循环通过相同的数据帧，当我结束第二个循环时，我想要索引，第一个循环的行从第二个循环结束的地方开始（即index = index2在下面的代码中）。这是我的示例代码：

    from pandas import *

    df = DataFrame(
        {'to': ['spot1','spot2','spot3','spot4','spot1','spot5'],'from': ['Denver','Denver','Cleveland','Timbuktu']})`

    for index,row in df.iterrows():     #iterate to find functions for measurement    
        
        value1 = row['from'] 
                
        for index2,row2 in islice(df.iterrows(),index+1,None):  #use islice to start on next row

            value2 = row2['from']

            if value1 != value2:            
                print('leaving from new destination)
                
                index = index2  #start outside loop at location where inside loop finished (ie value1 = Cleveland on second outside iteration)             
                
                break   #stop loop to prevent needless looping

            elif value1 == value2:
        
                print('still leaving from same destination)

解决方法

好吧，我想您只需要一个即可，例如：

import pandas as pd
df = pd.DataFrame(
  {
     'to': ['spot1','spot2','spot3','spot4','spot1','spot5'],'from':['Denver','Denver','Cleveland','Timbuktu']
  }
)
current_idx = df.iloc[0].name  # get first index
current_from = df.loc[current_idx,"from"]

for idx,row in df.iloc[1:].iterrows():  # Iter from the second row
  if current_from == row["from"]:
    print('still leaving from same destination')
  else:
     print('leaving from new destination')
     current_idx = idx
     current_from = row["from"]

老实说，您不需要保持索引hahah