convert list of lists in dataframe
Question:
I have a following data
0 [[-0.932, 2.443, -1....
1 [[-1.099, 2.140, -1.4...
2 [[-0.985, -1.654, -1....
3 [[-1.339, 2.070, -0....
4 [[-1.119, 2.788, -2....
...
494 [[-0.023, 2.688, -1...
495 [[1.897, 0.0, -2.249,...
496 [[1.538, 2.349, -0.6...
497 [[-0.141, 2.320, -0...
498 [[-0.483, 1.587, -1....
Length: 499, dtype: object
In each row are about 80 lists consisted (list of lists) and I would like to turn them into columns and to get the data:
ID col1 col2 ... col80
1.1.2020 0 -0.932 ...
2.1.2020 0 2.443 ...
3.1.2020 0 -1 ...
1.1.2020 1 -1.099
2.1.2020 1 2.140
3.1.2020 1 -1.4 ...
where the column ID is from the lists indicator (0,1,..,498). The index column (1.1.2020 2.1.2020..) is saved as another object (date). Is this possible and how?
Answers:
Possible. If that is the answer you are looking for. 🙂
Let’s say you had data like:
import numpy as np
import pandas as pd
ser = pd.Series(np.arange(90).reshape(10, 3, 3).tolist())
0 [[0, 1, 2], [3, 4, 5], [6, 7, 8]]
1 [[9, 10, 11], [12, 13, 14], [15, 16, 17]]
2 [[18, 19, 20], [21, 22, 23], [24, 25, 26]]
3 [[27, 28, 29], [30, 31, 32], [33, 34, 35]]
4 [[36, 37, 38], [39, 40, 41], [42, 43, 44]]
5 [[45, 46, 47], [48, 49, 50], [51, 52, 53]]
6 [[54, 55, 56], [57, 58, 59], [60, 61, 62]]
7 [[63, 64, 65], [66, 67, 68], [69, 70, 71]]
8 [[72, 73, 74], [75, 76, 77], [78, 79, 80]]
9 [[81, 82, 83], [84, 85, 86], [87, 88, 89]]
dtype: object
then I think you can do the bulk of the work like so:
out = ser.explode().apply(pd.Series).reset_index(names="ID")
ID 0 1 2
0 0 0 1 2
1 0 3 4 5
2 0 6 7 8
3 1 9 10 11
4 1 12 13 14
5 1 15 16 17
6 2 18 19 20
7 2 21 22 23
8 2 24 25 26
9 3 27 28 29
10 3 30 31 32
11 3 33 34 35
12 4 36 37 38
13 4 39 40 41
14 4 42 43 44
15 5 45 46 47
16 5 48 49 50
17 5 51 52 53
18 6 54 55 56
19 6 57 58 59
20 6 60 61 62
21 7 63 64 65
22 7 66 67 68
23 7 69 70 71
24 8 72 73 74
25 8 75 76 77
26 8 78 79 80
27 9 81 82 83
28 9 84 85 86
29 9 87 88 89
but you’ll need to rename the columns and change the index yourself (how are you determining those dates?)
I have a following data
0 [[-0.932, 2.443, -1....
1 [[-1.099, 2.140, -1.4...
2 [[-0.985, -1.654, -1....
3 [[-1.339, 2.070, -0....
4 [[-1.119, 2.788, -2....
...
494 [[-0.023, 2.688, -1...
495 [[1.897, 0.0, -2.249,...
496 [[1.538, 2.349, -0.6...
497 [[-0.141, 2.320, -0...
498 [[-0.483, 1.587, -1....
Length: 499, dtype: object
In each row are about 80 lists consisted (list of lists) and I would like to turn them into columns and to get the data:
ID col1 col2 ... col80
1.1.2020 0 -0.932 ...
2.1.2020 0 2.443 ...
3.1.2020 0 -1 ...
1.1.2020 1 -1.099
2.1.2020 1 2.140
3.1.2020 1 -1.4 ...
where the column ID is from the lists indicator (0,1,..,498). The index column (1.1.2020 2.1.2020..) is saved as another object (date). Is this possible and how?
Possible. If that is the answer you are looking for. 🙂
Let’s say you had data like:
import numpy as np
import pandas as pd
ser = pd.Series(np.arange(90).reshape(10, 3, 3).tolist())
0 [[0, 1, 2], [3, 4, 5], [6, 7, 8]]
1 [[9, 10, 11], [12, 13, 14], [15, 16, 17]]
2 [[18, 19, 20], [21, 22, 23], [24, 25, 26]]
3 [[27, 28, 29], [30, 31, 32], [33, 34, 35]]
4 [[36, 37, 38], [39, 40, 41], [42, 43, 44]]
5 [[45, 46, 47], [48, 49, 50], [51, 52, 53]]
6 [[54, 55, 56], [57, 58, 59], [60, 61, 62]]
7 [[63, 64, 65], [66, 67, 68], [69, 70, 71]]
8 [[72, 73, 74], [75, 76, 77], [78, 79, 80]]
9 [[81, 82, 83], [84, 85, 86], [87, 88, 89]]
dtype: object
then I think you can do the bulk of the work like so:
out = ser.explode().apply(pd.Series).reset_index(names="ID")
ID 0 1 2
0 0 0 1 2
1 0 3 4 5
2 0 6 7 8
3 1 9 10 11
4 1 12 13 14
5 1 15 16 17
6 2 18 19 20
7 2 21 22 23
8 2 24 25 26
9 3 27 28 29
10 3 30 31 32
11 3 33 34 35
12 4 36 37 38
13 4 39 40 41
14 4 42 43 44
15 5 45 46 47
16 5 48 49 50
17 5 51 52 53
18 6 54 55 56
19 6 57 58 59
20 6 60 61 62
21 7 63 64 65
22 7 66 67 68
23 7 69 70 71
24 8 72 73 74
25 8 75 76 77
26 8 78 79 80
27 9 81 82 83
28 9 84 85 86
29 9 87 88 89
but you’ll need to rename the columns and change the index yourself (how are you determining those dates?)