Suppose I have the following list:
m=[1,2,[1],1,2,[1]]
I wish to take away all duplicates. If it were not for the brackets inside the the list, then I could use:
m=list(set(m))
but when I do this, I get the error:
unhashable type 'set'.
What command will help me remove duplicates so that I could only be left with the list
m=[1,2,[1]]
Thank you
You can do something along these lines:
m=[1,2,[1],1,2,[1]]
seen=set()
nm=[]
for e in m:
try:
x={e}
x=e
except TypeError:
x=frozenset(e)
if x not in seen:
seen.add(x)
nm.append(e)
>>> nm
[1, 2, [1]]
From comments: This method preserves the order of the original list. If you want the numeric types in order first and the other types second, you can do:
sorted(nm, key=lambda e: 0 if isinstance(e, (int,float)) else 1)
result = [] for i in m: flag = True for j in m: if i == j: flag = False if flag: result.append(i)
Result will be: [1,2,[1]]
There are ways to make this code shorter, but I'm writing it more verbosely for readability. Also, note that this method is O(n^2), so I wouldn't recommend for long lists. But benefits is the simplicity.
The first step will be to convert the inner lists to tuples:
>> new_list = [tuple(i) if type(i) == list else i for i in m]
Then create a set to remove duplicates:
>> no_duplicates = set(new_list)
>> no_duplicates
{1, 2, (1,)}
and you can convert that into list if you wish.
For a more generic solution you can serialize each list item with pickle.dumps
before passing them to set()
, and then de-serialize the items with pickle.loads
:
import pickle
m = list(map(pickle.loads, set(map(pickle.dumps, m))))
If you want the original order to be maintained, you can use a dict (which has become ordered since Python 3.6+) instead of a set:
import pickle
m = list(map(pickle.loads, {k: 1 for k in map(pickle.dumps, m)}))
Or if you need to be compatible with Python 3.5 or earlier versions, you can use collections.OrderedDict
instead:
import pickle
from collections import OrderedDict
m = list(map(pickle.loads, OrderedDict((k, 1) for k in map(pickle.dumps, m))))
Simple Solution,
m=[1,2,[1],1,2,[1]]
l= []
for i in m:
if i not in l:
l.append(i)
print(l)
[1, 2, [1]]
[Program finished]
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