Pandas filter with month
Question:
I have a dataframe that looks like below
Date 3tier1 2tier1 2tier2 2tier3
2001-12-01 08:00:00+08:00 64.2946632 47.25716228 17.04223722 64.2946632
2001-12-02 08:00:00+08:00 63.87557277 47.28039931 16.59856272 63.87557277
2001-12-03 08:00:00+08:00 63.43537544 47.29749771 16.14010007 63.43537544
2001-12-04 08:00:00+08:00 62.97632974 47.30827399 15.66932265 62.97632974
2001-12-05 08:00:00+08:00 62.50030363 47.31253544 15.18832366 62.50030363
2001-12-06 08:00:00+08:00 62.0088254 47.31007936 14.6988683 62.0088254
2001-12-07 08:00:00+08:00 61.50312807 47.30069244 14.20243945 61.50312807
2001-12-08 08:00:00+08:00 60.98418775 47.28415013 13.70027744 60.98418775
2001-12-09 08:00:00+08:00 60.45275675 47.26021591 13.19341477 60.45275675
2002-05-12 08:00:00+08:00 60.06028069 45.43305289 14.67607992 60.06028069
2002-05-13 08:00:00+08:00 60.89978183 45.34347401 15.60584812 60.89978183
2002-05-14 08:00:00+08:00 61.68801293 45.24973017 16.48786436 61.68801293
2002-05-15 08:00:00+08:00 62.42284657 45.15191829 17.31994301 62.42284657
2002-05-16 08:00:00+08:00 63.10251078 45.05013361 18.10025949 63.10251078
2002-05-17 08:00:00+08:00 63.72553989 44.94447026 18.82729998 63.72553989
2002-05-18 08:00:00+08:00 64.29072138 44.83502163 19.49980764 64.29072138
2002-05-19 08:00:00+08:00 64.79704208 44.7218804 20.11672858 64.79704208
2002-06-03 08:00:00+08:00 64.79007123 42.64210728 22.14835854 64.79007123
2002-06-04 08:00:00+08:00 64.24049486 42.48172309 21.76076277 64.24049486
2002-06-05 08:00:00+08:00 63.61590394 42.31904254 21.30190026 63.61590394
The above data is actually filtered from another dataframe using the below code
df = df[df['3tier1'].between(60, 65)]
I wanted to filter values between 60 and 65 using column 3tier1 and used the above code.
Now, i want a filter to pick up data only for the months May and June i.e month 5 and 6 from the date column from the above filtered data. Is there a pandas function that can do this easily. Date column contains time as well 08:00:00+08:00 , this is basically 8 am in timezone +8
Answers:
To filter on month, you can pass the column series to pd.to_datetime then access month by dt.month then use isin method and pass list of months you are interested in i.e. 5 and 6:
df[df['3tier1'].between(60, 65) & pd.to_datetime(df['Date']).dt.month.isin([5,6])]
Date 3tier1 2tier1 2tier2 2tier3
9 2002-05-12 08:00:00+08:00 60.060281 45.433053 14.676080 60.060281
10 2002-05-13 08:00:00+08:00 60.899782 45.343474 15.605848 60.899782
11 2002-05-14 08:00:00+08:00 61.688013 45.249730 16.487864 61.688013
12 2002-05-15 08:00:00+08:00 62.422847 45.151918 17.319943 62.422847
13 2002-05-16 08:00:00+08:00 63.102511 45.050134 18.100259 63.102511
14 2002-05-17 08:00:00+08:00 63.725540 44.944470 18.827300 63.725540
15 2002-05-18 08:00:00+08:00 64.290721 44.835022 19.499808 64.290721
16 2002-05-19 08:00:00+08:00 64.797042 44.721880 20.116729 64.797042
17 2002-06-03 08:00:00+08:00 64.790071 42.642107 22.148359 64.790071
18 2002-06-04 08:00:00+08:00 64.240495 42.481723 21.760763 64.240495
19 2002-06-05 08:00:00+08:00 63.615904 42.319043 21.301900 63.615904
I have a dataframe that looks like below
Date 3tier1 2tier1 2tier2 2tier3
2001-12-01 08:00:00+08:00 64.2946632 47.25716228 17.04223722 64.2946632
2001-12-02 08:00:00+08:00 63.87557277 47.28039931 16.59856272 63.87557277
2001-12-03 08:00:00+08:00 63.43537544 47.29749771 16.14010007 63.43537544
2001-12-04 08:00:00+08:00 62.97632974 47.30827399 15.66932265 62.97632974
2001-12-05 08:00:00+08:00 62.50030363 47.31253544 15.18832366 62.50030363
2001-12-06 08:00:00+08:00 62.0088254 47.31007936 14.6988683 62.0088254
2001-12-07 08:00:00+08:00 61.50312807 47.30069244 14.20243945 61.50312807
2001-12-08 08:00:00+08:00 60.98418775 47.28415013 13.70027744 60.98418775
2001-12-09 08:00:00+08:00 60.45275675 47.26021591 13.19341477 60.45275675
2002-05-12 08:00:00+08:00 60.06028069 45.43305289 14.67607992 60.06028069
2002-05-13 08:00:00+08:00 60.89978183 45.34347401 15.60584812 60.89978183
2002-05-14 08:00:00+08:00 61.68801293 45.24973017 16.48786436 61.68801293
2002-05-15 08:00:00+08:00 62.42284657 45.15191829 17.31994301 62.42284657
2002-05-16 08:00:00+08:00 63.10251078 45.05013361 18.10025949 63.10251078
2002-05-17 08:00:00+08:00 63.72553989 44.94447026 18.82729998 63.72553989
2002-05-18 08:00:00+08:00 64.29072138 44.83502163 19.49980764 64.29072138
2002-05-19 08:00:00+08:00 64.79704208 44.7218804 20.11672858 64.79704208
2002-06-03 08:00:00+08:00 64.79007123 42.64210728 22.14835854 64.79007123
2002-06-04 08:00:00+08:00 64.24049486 42.48172309 21.76076277 64.24049486
2002-06-05 08:00:00+08:00 63.61590394 42.31904254 21.30190026 63.61590394
The above data is actually filtered from another dataframe using the below code
df = df[df['3tier1'].between(60, 65)]
I wanted to filter values between 60 and 65 using column 3tier1 and used the above code.
Now, i want a filter to pick up data only for the months May and June i.e month 5 and 6 from the date column from the above filtered data. Is there a pandas function that can do this easily. Date column contains time as well 08:00:00+08:00 , this is basically 8 am in timezone +8
To filter on month, you can pass the column series to pd.to_datetime then access month by dt.month then use isin method and pass list of months you are interested in i.e. 5 and 6:
df[df['3tier1'].between(60, 65) & pd.to_datetime(df['Date']).dt.month.isin([5,6])]
Date 3tier1 2tier1 2tier2 2tier3
9 2002-05-12 08:00:00+08:00 60.060281 45.433053 14.676080 60.060281
10 2002-05-13 08:00:00+08:00 60.899782 45.343474 15.605848 60.899782
11 2002-05-14 08:00:00+08:00 61.688013 45.249730 16.487864 61.688013
12 2002-05-15 08:00:00+08:00 62.422847 45.151918 17.319943 62.422847
13 2002-05-16 08:00:00+08:00 63.102511 45.050134 18.100259 63.102511
14 2002-05-17 08:00:00+08:00 63.725540 44.944470 18.827300 63.725540
15 2002-05-18 08:00:00+08:00 64.290721 44.835022 19.499808 64.290721
16 2002-05-19 08:00:00+08:00 64.797042 44.721880 20.116729 64.797042
17 2002-06-03 08:00:00+08:00 64.790071 42.642107 22.148359 64.790071
18 2002-06-04 08:00:00+08:00 64.240495 42.481723 21.760763 64.240495
19 2002-06-05 08:00:00+08:00 63.615904 42.319043 21.301900 63.615904