jean - 1 year ago 117

Python Question

I have a python list as follows:

`my_list =`

[[25, 1, 0.65],

[25, 3, 0.63],

[25, 2, 0.62],

[50, 3, 0.65],

[50, 2, 0.63],

[50, 1, 0.62]]

I want to order them according to this rule:

`1 --> [0.65, 0.62] <--25, 50`

2 --> [0.62, 0.63] <--25, 50

3 --> [0.63, 0.65] <--25, 50

So the expected result is as follows:

`Result = [[0.65, 0.62],[0.62, 0.63],[0.63, 0.65]]`

How to do it guys?

I tried as follows:

`df = pd.DataFrame(my_list,columns=['a','b','c'])`

res = df.groupby(['b', 'c']).get_group('c')

print res

ValueError: must supply a tuple to get_group with multiple grouping keys

Answer Source

You can sort your list with native python, but I find it easiest to get your required list using numpy. Since you were going to use pandas anyway, I consider this to be an acceptable solution:

```
from operator import itemgetter
import numpy as np
# or just use pandas.np if you have that already imported
my_list = [[25, 1, 0.65],
[25, 3, 0.63],
[25, 2, 0.62],
[50, 3, 0.65],
[50, 2, 0.63],
[50, 1, 0.62]]
sorted_list = sorted(my_list,key=itemgetter(1,0)) # sort by second and first column
sliced_array = np.array(sorted_list)[:,-1].reshape(-1,2)
final_list = sliced_array.tolist() # to get a list
```

The main point is to use `itemgetter`

to sort your list on two columns one after the other. The resulting sorted list contains the required elements in its third column, which I extract with numpy. It could be done with native python, but if you're already using numpy/pandas, this should be natural.