nicolax9777 :
I'm new in python. I have a large dataframe as like that :
ID x y
0 1 x1 y1
1 0 x2 y2
2 0 x3 y3
3 2 x4 y4
4 1 x5 y5
5 2 x6 y6
I would like to take couples of (x;y) between the IDs 1 and 2, in a dataframe like this :
coordinates
0 (x1,y1), (x2,y2), (x3,y3), (x4,y4)
1 (x5,y5), (x6,y6)
I already tried a double for iteration but it is to long to compute. How can I get this thing?
jezrael :
One idea is create groups by each 1 starting values and aggregate custom lambda function for tuples:
df['new'] = (df['ID'] == 1).cumsum()
print (df)
ID x y new
0 1 x1 y1 1
1 0 x2 y2 1
2 0 x3 y3 1
3 2 x4 y4 1
4 1 x5 y5 2
5 2 x6 y6 2
df1 = (df.groupby('new')['x','y']
.apply(lambda x: list(map(tuple, x.values.tolist())))
.reset_index(name='coordinates'))
print (df1)
new coordinates
0 1 [(x1, y1), (x2, y2), (x3, y3), (x4, y4)]
1 2 [(x5, y5), (x6, y6)]
Similar solution with no new column:
df1 = (df.groupby((df['ID'].rename('new') == 1).cumsum())['x','y']
.apply(lambda x: list(map(tuple, x.values.tolist())))
.reset_index(name='coordinates'))
print (df1)
new coordinates
0 1 [(x1, y1), (x2, y2), (x3, y3), (x4, y4)]
1 2 [(x5, y5), (x6, y6)]
EDIT:
print (df)
ID x y
0 1 x1 y1
1 0 x2 y2
2 0 x3 y3
3 2 x4 y4
4 0 x7 y7
4 0 x8 y8
4 1 x5 y5
5 2 x6 y6
g = df['ID'].eq(1).cumsum()
s = df['ID'].shift().eq(2).cumsum()
df = df[s.groupby(g).transform('min').eq(s)]
print (df)
ID x y
0 1 x1 y1
1 0 x2 y2
2 0 x3 y3
3 2 x4 y4
4 1 x5 y5
5 2 x6 y6
df1 = (df.groupby((df['ID'].rename('new') == 1).cumsum())['x','y']
.apply(lambda x: list(map(tuple, x.values.tolist())))
.reset_index(name='coordinates'))
print (df1)
new coordinates
0 1 [(x1, y1), (x2, y2), (x3, y3), (x4, y4)]
1 2 [(x5, y5), (x6, y6)]