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Basically, I'm trying to do remove any lists that begin with the same value. For example, two of the below begin with the number 1:

a = [[1,2],[1,0],[2,4],[3,5]]

Because the value 1 exists at the start of two of the lists -- I need to remove both so that the new list becomes:

b = [[2,4],[3,5]]

How can I do this?

I've tried the below, but the output is: [[1, 2], [2, 4], [3, 5]]

def unique_by_first_n(n, coll):
 seen = set()
 for item in coll:
 compare = tuple(item[:n])
 print compare # Keep only the first `n` elements in the set
 if compare not in seen:
 seen.add(compare)
 yield item

a = [[1,2],[1,0],[2,4],[3,5]]

filtered_list = list(unique_by_first_n(1, a))

You can use collections.Counter with list comprehension to get sublists whose first item appears only once:

from collections import Counter
c = Counter(n for n, _ in a)
b = [[x, y] for x, y in a if c[x] == 1]

An efficient solution would be to create a Counter object to hold the occurrences of the first elements, and then filter the sub-lists in the main list:

from collections import Counter
counts = Counter(l[0] for l in a)
filtered = [l for l in a if counts[l[0]] == 1]
#[[2, 4], [3, 5]]

If you are happy to use a 3rd party library, you can use Pandas:

import pandas as pd

a = [[1,2],[1,0],[2,4],[3,5]]

df = pd.DataFrame(a)
b = df.drop_duplicates(subset=[0], keep=False).values.tolist()

print(b)

[[2, 4], [3, 5]]

The trick is the keep=False argument, described in the docs for pd.DataFrame.drop_duplicates.

Solution 1

a = [[1,2],[1,0],[2,4],[3,5]]
b = 
for item in a:
 i = 0
 if item[0] == a[i][0]:
 i =+ 1
 continue
 else:
 b.append(item)
 i += 1

Solution 2

a = [[1,2],[1,0],[2,4],[3,5]]
b = 

for item in a:
 for i in range(0, len(a)):
 if item[0] == a[i][0]:
 break
 else:
 if item in b:
 continue
 else:
 b.append(item)

Output

(xenial)vash@localhost:~/pcc/10$ python3 remove_help.py 
[[1, 2], [1, 0], [2, 4], [3, 5]]
[[2, 4], [3, 5]]

Achieved your goal no complex methods involed!
Enjoy!