Python: Create sublist based on values/conditions

Question

As a result of a linear program I get a nested list out of a dict which looks like this:

lst = [['a0', 1, 'x_01', 'p_01', 39],
       ['a0', 1, 'x_01', 'p_02', 19],
       ['a0', 1, 'x_02', 'p_01', 10],
       ['a0', 1, 'x_02', 'p_02', 5],
       ['a1', 1, 'x_01', 'y_01', 'p_01', 5],
       ['a1', 1, 'x_01', 'y_01', 'p_02', 5],
       ['a1', 1, 'x_02', 'y_01', 'p_02', 5],
       ['a1', 1, 'x_02', 'y_01', 'p_02', 5]]

Any kind of 'a' indicates a variable which was solved for a range of sets (eg. 1, 'x_01', 'p_01' - etc.) with a value represented by the last element of the list.

Now the task is to create sub lists only containing lists of the same variable. (In the next step they will be passed to a pd.df) The list of lists could look like this or something similar:

new_lst = [[['a0', 1, 'x_01', 'p_01', 39],
       ['a0', 1, 'x_01', 'p_02', 19],
       ['a0', 1, 'x_02', 'p_01', 10],
       ['a0', 1, 'x_02', 'p_02', 5]],
       [['a1', 1, 'x_01', 'y_01', 'p_01', 5],
       ['a1', 1, 'x_01', 'y_01', 'p_02', 5],
       ['a1', 1, 'x_02', 'y_01', 'p_02', 5],
       ['a1', 1, 'x_02', 'y_01', 'p_02', 5]]]

What I've tried so far is:

st0 = list(set([x[0] for x in lst]))

to get all the unique variables.

And this, but it obviously does not do the job:

n = [x for x in lst for i in st0 if x[0]==i]

What I try here is based on something like this:

n1 = [x for x in lst if x[0]==st0[0]]
n2 = [x for x in lst if x[0]==st0[1]]
ii = [n1,n2]

Based on the model size I have 10, 20 or more different variables of different dimensions. Also the models need to be run several times to test different parameters.

Hard coding is possible, but not desired. If anyone has a good idea please feel free to help me out on this one.

Austin · Accepted Answer · 2019-02-16 16:34:08Z

4

Use itertools.groupby:

from itertools import groupby

lst = [['a0', 1, 'x_01', 'p_01', 39],
       ['a0', 1, 'x_01', 'p_02', 19],
       ['a0', 1, 'x_02', 'p_01', 10],
       ['a0', 1, 'x_02', 'p_02', 5],
       ['a1', 1, 'x_01', 'y_01', 'p_01', 5],
       ['a1', 1, 'x_01', 'y_01', 'p_02', 5],
       ['a1', 1, 'x_02', 'y_01', 'p_02', 5],
       ['a1', 1, 'x_02', 'y_01', 'p_02', 5]]

new_lst = []
s_lst = sorted(lst, key=lambda x: x[0])
for k, v in groupby(s_lst, key=lambda x: x[0]):
    new_lst.append(list(v))

print(new_lst)

Output:

[[['a0', 1, 'x_01', 'p_01', 39], 
  ['a0', 1, 'x_01', 'p_02', 19], 
  ['a0', 1, 'x_02', 'p_01', 10], 
  ['a0', 1, 'x_02', 'p_02', 5]], 
 [['a1', 1, 'x_01', 'y_01', 'p_01', 5], 
  ['a1', 1, 'x_01', 'y_01', 'p_02', 5], 
  ['a1', 1, 'x_02', 'y_01', 'p_02', 5], 
  ['a1', 1, 'x_02', 'y_01', 'p_02', 5]]]

answered Feb 16, 2019 at 16:34

Austin

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4 Comments

JGFMK Over a year ago

Well you had a keen eye Austin. It was hard to see wood from the trees. The extra square brackets.... Had no idea what op wanted. That was all that was needed to hone in on actual need. Some sort of inverse flatten.

Austin Over a year ago

@JGFMK, still I'm not sure if this is OP wanted. Waiting for his response. :)

AvePazuzu Over a year ago

@Austin: This did the job perfectly well. I will try it on a lager data set to test performance on Monday. Tanks!

AvePazuzu Over a year ago

@JGFMK: I Thought some more information might help, but you are right, just some extra square brackets.

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