How to delete certain number from a list in list using the index in python?-CodePudding

I have a list in a list, and I am trying to delete the third number of each sublist, but every time I am getting an error TypeError: list indices must be integers or slices, not list

a = [[0.0, 0.0, 0.0], [0.19, 0.36, 0.0], [0.24, 0.42, 0.0], [0.16, 0.08, 0.0], [0.05, -0.57, 0.0] ]

Desired result:-

a_updated = [[0.0, 0.0], [0.19, 0.36], [0.24, 0.42], [0.16, 0.08], [0.05, -0.57] ]

In the second part of my code, I wanted to merge this sublist according to a dictionary shown below, for example, the first value of dictionary:- 1: [1, 2] shows the merging of 1st and 2nd values i.e. [0, 0, 0.19, 0.36].

I guess this part of my code is right!

dict_a = { {1: [1, 2], 2: [2, 4], 3: [3, 5], 4: [4, 5] }

my attempt:-

dict_a = { 1: [1, 2], 2: [2, 4], 3: [3, 5], 4: [4, 5]}

a = [[0.0, 0.0], [0.19, 0.36], [0.24, 0.42], [0.16, 0.08], [0.05, -0.57]]
       

# first part 
for i in a:
    for j in a[i]:
            del j[2]
    print(j)
    
    
#second part   
a_list = []
list_of_index = []
for i in dict_a:
    index= []
    a_list.append(index)
    for j in dict_a_updated[i]:
            print(j-1)
            index.extend(a_updated[j-1])    
    print('index',index)

Error output -


file "D:\python programming\random python files\4 may axial dis.py", line 18, in <module>
    for j in X[i]:

TypeError: list indices must be integers or slices, not list

CodePudding user response：

You can slice the sublists in a list comprehension to build a_updated:

a_updated = [s_lst[:2] for s_lst in a]

Output:

[[0.0, 0.0], [0.19, 0.36], [0.24, 0.42], [0.16, 0.08], [0.05, -0.57]]

To build dict_a_updated, you can use a loop. Note that list indices start from 0 in Python but your index starts from 1, so we have to subtract 1 here:

dict_a_updated = {}
for k, v in dict_a.items():
    tmp = []
    for i in v:
        tmp.extend(a_updated[i-1])
    dict_a_updated[k] = tmp

Output:

{1: [0.0, 0.0, 0.19, 0.36],
 2: [0.19, 0.36, 0.16, 0.08],
 3: [0.24, 0.42, 0.05, -0.57],
 4: [0.16, 0.08, 0.05, -0.57]}

CodePudding user response：

Given a list of lists:

a = [[0.0, 0.0, 0.0], [0.19, 0.36, 0.0], [0.24, 0.42, 0.0], [0.16, 0.08, 0.0], [0.05, -0.57, 0.0]]

The first solution: if the third element is the last element in all arrays.

firstSolution = [el[:-1] for el in a]
print(firstSolution)

The second solution: is to remove the element by its index.

for el in a:
  el.pop(2)
print(a)

CodePudding user response：

You can achieve you goal with a single dictionary comprehension and itertools.chain, without needing to first rework a:

from itertools import chain
out = {k: list(chain.from_iterable(a[i-1][:2] for i in v))
       for k,v in dict_a.items()}

Output:

{1: [0.0, 0.0, 0.19, 0.36],
 2: [0.19, 0.36, 0.16, 0.08],
 3: [0.24, 0.42, 0.05, -0.57],
 4: [0.16, 0.08, 0.05, -0.57]}

CodePudding user response：

a = [[0.0, 0.0, 0.0], [0.19, 0.36, 0.0], [0.24, 0.42, 0.0], [0.16, 0.08, 0.0], [0.05, -0.57, 0.0] ]

dict_a = {1: [1, 2], 2: [2, 4], 3: [3, 5], 4: [4, 5] }
    
# first part 
for i in range(0,len(a)):  # for every sublist position
    a[i] = a[i][0:2]
    
dict_lists = {}
for key,value in dict_a.items():
    dict_lists[key] = [a[value[0]-1], a[value[1]-1]]

Output:

In[19]: dict_lists
Out[19]: 
{1: [[0.0, 0.0], [0.19, 0.36]],
 2: [[0.19, 0.36], [0.16, 0.08]],
 3: [[0.24, 0.42], [0.05, -0.57]],
 4: [[0.16, 0.08], [0.05, -0.57]]}