What problems are you running into with apply? It works for this toy example here and the group lengths are different: In [82]: df Out[82]: X Y 0 0 -0.631214 1 0 0.783142 2 0 0.526045 3 1 -1.750058 4 1 1.163868 5 1 1.625538 6 1 0.076105 7 2 0.183492 8 2 0.541400 9 … Read more
Because the indexes are not exactly equal, NaNs will result. Either one or both of the indexes must be changed to match. Example: mydata = mydata.set_index(DWDATA.index) The above will change the index of the ‘mydata’ DataFrame to match the index of the ‘DWDATA’ DataFrame. Since the number of rows are exactly equal for the two … Read more
Starting from Pandas 0.25.0, there is internal method DataFrame.explode(), which was designed just for that: res = df.explode(“b”) output In [98]: res Out[98]: a b 0 1 1 0 1 2 1 2 2 1 2 3 1 2 4 2 3 5 Solution for Pandas versions < 0.25: generic vectorized approach – will work … Read more
As mentioned in the comments, you have a dupe column:
Try using xs to be very precise: In [5]: df.xs(‘a’, level=0) Out[5]: value1 value2 group2 c 1.1 7.1 c 2.0 8.0 d 3.0 9.0 In [6]: df.xs(‘c’, level=”group2″) Out[6]: value1 value2 group1 a 1.1 7.1 a 2.0 8.0
As @Hatshepsut pointed out in the comments, from_items is deprecated as of version 0.23. The link suggests to use from_dict instead, so the old answer can be modified to: pd.DataFrame.from_dict(dict(zip(s.index, s.values))) ————————————————–OLD ANSWER————————————————————- You can use from_items like this (assuming that your lists are of the same length): pd.DataFrame.from_items(zip(s.index, s.values)) 0 1 0 1 4 … Read more
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