I have a dataframe :
start end
1 10
26 50
6 15
1 5
11 25
I expect following dataframe :
start end
1 10
11 25
26 50
1 5
6 15
here sort order is noting but end of nth row must be start+1 of n+1th row.If not found, search for other starts where start is one.
can anyone suggest what combination of sort and group by can I use to convert above dataframe in required format?
You could transform the df to a list and then do:
l=[1,10,26,50,6,15,1,5,11,25]
result=[]
for x in range(int(len(l)/2)):
result.append(sorted([l[2*x],l[2*x+1]])[1])
result.append(sorted([l[2*x],l[2*x+1]])[0])
This will give you result:
[1, 10, 26, 50, 6, 15, 1, 5, 11, 25]
To transform the original df to list you can do:
startcollist=df['start'].values.tolist()
endcollist=df['end'].values.tolist()
l=[]
for index, each in enumerate(originaldf):
l.append(each)
l.append(endcollist[index])
You can then transform result back to a dataframe:
df=pd.DataFrame({'start':result[1::2], 'end':result[0::2]})
Giving the result:
end start
0 10 1
1 50 26
2 15 6
3 5 1
4 25 11
The expression result[1::2] gives every odd element of result, result[0::2] gives every even element. For explanation, see here: https://stackoverflow.com/a/12433705/8565438
