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 [pandas] How to reshape the list
I have blow list, I can reshape it but I want to reshape based on other condition:

import pandas as pd
import numpy as np
lst = ['A','',3,'A','',0,'A',3,2]
aa = np.reshape(lst,(-1,3))
aa = pd.DataFrame(aa, columns=['A','B','C'])
but I want to break the sub-list at each 'A'
A		3
A		0
A	3	2
I want to break at 'A' because some time columns may change,
if I have a list like:

lst = ['A','',3,'4','A','',0,'2020/07/25 10:13:23','A',3,2,''2020/07/25 10:13:29']
in this case, i need to change np.reshape(lst,(-1,4)), which I want to avoide
You can use np.split and np.where to do this
np.vstack(np.split(lst, np.where(np.array(lst) == 'A')[0][1:]))[:,1:]
array([['', '3', '4'], ['', '0', '2020/07/25 10:13:23'], ['3', '2', '2020/07/25 10:13:29']], dtype='<U19')
Though it splits at "A", but I want to keep "A" as my first column row element.
'A','',0,'2020/07/25 10:13:23'
'A',3,2,''2020/07/25 10:13:29'
(Jul-25-2020, 05:25 AM)Mekala Wrote: Though it splits at "A", but I want to keep "A" as my first column row element.
Just remove [:,1:] from the previous solution.
what is function of [0][1:])? may I learn.
np.where returns a tuple. The first item of this tuple is an array of indecies where A is occurred. So, we need to pass np.where(...)[0] to the np.split function. However, the first element of lst is 'A' and we don't need to split array at this position. So, we just excluded the index of first occurrence of A from splittng [1:] -- means select all indecies but not the first (first is 0).
Thus, [0] -- selects the first element of the tuple that is an array of indecies where A is occurred. and [1:] selects all such indecies except the first.
Mekala likes this post
OK, Thanks.

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