I'm trying to use an if to verify if a value in a dataframe equals a certain value i got before. Here's an extract of the dataframe: usager heures demandes heure arrivée ascenseur heures arrivée étage \ 0 1181.0 2019-04-27 07:00:27 2019-04-27 07:00:27 2019-04-27 07:00:43 1 1181.0 2019-04-27 12:11:54 2019-04-27 12:12:10 2019-04-27 12:12:26 2 1181.0 2019-04-27 12:55:52 2019-04-27 12:56:06 2019-04-27 12:56:22 3 1181.0 2019-04-27 17:25:49 2019-04-27 17:26:05 2019-04-27 17:26:21 0 826.0 2019-04-27 07:00:50 2019-04-27 07:00:50 2019-04-27 07:00:59 .. ... ... ... ... 3 677.0 2019-04-27 17:38:47 2019-04-27 17:39:17 2019-04-27 17:39:34 0 667.0 2019-04-27 09:59:38 2019-04-27 09:59:54 2019-04-27 10:00:09 1 667.0 2019-04-27 12:23:24 2019-04-27 12:23:39 2019-04-27 12:23:54 2 667.0 2019-04-27 13:25:16 2019-04-27 13:25:28 2019-04-27 13:25:43 3 667.0 2019-04-27 18:01:03 2019-04-27 18:01:18 2019-04-27 18:01:33 mode transport etage_demandes attente temps dans ascenseur anomalie 0 Ascenseur 33.0 0.0 16.0 False 1 Ascenseur 0.0 16.0 16.0 False 2 Ascenseur 33.0 14.0 16.0 False 3 Ascenseur 0.0 16.0 16.0 False 0 Ascenseur 18.0 0.0 9.0 False .. ... ... ... ... ... 3 Ascenseur 0.0 30.0 17.0 False 0 Ascenseur 30.0 16.0 15.0 False 1 Ascenseur 0.0 15.0 15.0 False 2 Ascenseur 30.0 12.0 15.0 False 3 Ascenseur 0.0 15.0 15.0 False With 'usager' being an unique id for each person, i'm trying to find the level where each person works which is shown in etage_demandes when it's not null. Each row in the dataframe is a movement from a level to another and each person moves 4 times. Because of how the dataframe is made I tried to do this: memory=0 print(df['usager'][1:2]) etage_travail=[df[:0]['etage_demandes']] for i in range (4,len(df['usager'])): if (df['usager'][i-1:i]=!memory): memory=df['usager'][i-1:i] for i in range (0,4): etage_travail.append(df[i-1:i]["etage_demandes"]) Where I go through each row and get the work level for each person when the id is not the same as the id of the row before. but I can't seem to write the condition in a way so that it works. Continue reading...