I would like to create a new column where the values are the sum of the last 14 values of column atr1, How can I do it?
I tried
col = a.columns.get_loc('atr1')
a['atrsum'] = a.iloc[-14:,col].sum()
But I get only a fixed value in the new column. Dataframe below as reference.
time open high low close volume atr1
0 1620518400000 1.6206 1.8330 1.5726 1.7663 8.830913e+08 NaN
1 1620604800000 1.7662 1.8243 1.5170 1.6423 7.123049e+08 0.3073
2 1620691200000 1.6418 1.7791 1.5954 1.7632 5.243267e+08 0.1837
3 1620777600000 1.7633 1.8210 1.5462 1.5694 5.997101e+08 0.2748
4 1620864000000 1.5669 1.9719 1.5000 1.9296 1.567655e+09 0.4719
... ... ... ... ... ... ... ...
360 1651622400000 0.7712 0.8992 0.7677 0.8985 2.566498e+08 0.1315
361 1651708800000 0.8986 0.9058 0.7716 0.7884 3.649706e+08 0.1342
362 1651795200000 0.7884 0.7997 0.7625 0.7832 2.440587e+08 0.0372
363 1651881600000 0.7832 0.7858 0.7467 0.7604 1.268089e+08 0.0391
364 1651968000000 0.7605 0.7663 0.7254 0.7403 1.751395e+08 0.0409
>Solution :
use expanding function of pandas
14 is the number of days, followed by the column you like to sum
a.expanding(14)['atr1'].sum()
I must be missing something in the question, ,my apologies.
I just used the data you shared and used the 2 previous days and this is the result
df['atrsum'] = df['atr1'].expanding(2).sum()
id time open high low close volume atr1 atrsum
0 0 1620518400000 1.6206 1.8330 1.5726 1.7663 8.830913e+08 NaN NaN
1 1 1620604800000 1.7662 1.8243 1.5170 1.6423 7.123049e+08 0.3073 NaN
2 2 1620691200000 1.6418 1.7791 1.5954 1.7632 5.243267e+08 0.1837 0.4910
3 3 1620777600000 1.7633 1.8210 1.5462 1.5694 5.997101e+08 0.2748 0.7658
4 4 1620864000000 1.5669 1.9719 1.5000 1.9296 1.567655e+09 0.4719 1.2377
5 360 1651622400000 0.7712 0.8992 0.7677 0.8985 2.566498e+08 0.1315 1.3692
6 361 1651708800000 0.8986 0.9058 0.7716 0.7884 3.649706e+08 0.1342 1.5034
7 362 1651795200000 0.7884 0.7997 0.7625 0.7832 2.440587e+08 0.0372 1.5406
8 363 1651881600000 0.7832 0.7858 0.7467 0.7604 1.268089e+08 0.0391 1.5797
9 364 1651968000000 0.7605 0.7663 0.7254 0.7403 1.751395e+08 0.0409 1.6206