Python: Test for equality of certain index in nested lists

user_5479213 0
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I have the following nested list:

nlist = [[2, 0, 4], [2, 0, 4], [0, 0, 4], [0, 0, 4]]

I would like to evaluate whether the first element of each list is equal. I have found methods to evaluate whether the entire nested lists are equal, but not specific indexes?

nested-lists python
4 Answers

You could make a set of the first elements of each sub list, and since there is no repetition in sets, you can see if the length of the set is one:

nlist = [[2, 0, 4], [2, 0, 4], [0, 0, 4], [0, 0, 4]]

result = len(set(l[0] for l in nlist)) == 1
print(result) # => False
user_62751030

I would use the all() function with a generator expression as its parameter.

all(sublist[0] == nlist[0][0] for sublist in nlist)

In short, all() will return True if every sublist[0] == nlist[0][0] for every sublist in the original list. It will return False if any item is not equal to nlist[0][0].

Rob_Adams_87470

you can find all the unique first element and then check index :

>>> from itertools import groupby
>>> nlist = [[2, 0, 4], [2, 0, 4], [0, 0, 4], [0, 0, 4]]
>>> c= [i for i,k in groupby([n[0] for n in nlist])]
>>> c
[2, 0]
>>> d=[[i for i, x in enumerate(nlist) if x[0]==j] for j in c]
>>> d
[[0, 1], [2, 3]]
>>> R=dict(zip(c, d))
>>> R
{0: [2, 3], 2: [0, 1]}

you get a dict with each first possible value and a list of index from your initial list.

user_60481140 
first_num = nlist[0][0]
for sublist in nlist:
    if first_num != sublist[0]
        print(False)
print(True)
user_75287310
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