Python: Remove values that are duplicated in the previous list inside a nested list-CodePudding

Example of a given nested list:

nList = [[2,5,99,99],[-3,8,1,2,10],[1, 7,100,10]]

Expectations: Remove values that are duplicated in the previous list.

Expected Output:

oList = [[2, 5, 99], [-3, 8, 1, 10], [7, 100]]

My Codes:

def RemoveDup(nList):
    lst = []
    for i in nList:
        for j in i:
            if j not in lst:
                lst.append(j)            
    return lst 

>>> print(RemoveDup([[2,5,99,99],[-3,8,1,2,10],[1, 7,100,10]]))
>>> [2, 5, 99, -3, 8, 1, 10, 7, 100]

I still couldn't figure out how to make the output a nested list like the expected output, any help and advice are appreciated!

CodePudding user response：

As there is no "previous" element to compare against the first element in the nested list then it must be dealt with as is.

This may achieve the real objective:

nList = [[2,5,99,99],[-3,8,1,2,10],[1, 7,100,10]]

output = [nList[0]]

for e in nList[1:]:
    t = []
    for x in e:
        if x not in output[-1]:
            t.append(x)
    output.append(t)

print(output)

Output:

[[2, 5, 99, 99], [-3, 8, 1, 10], [7, 100]]

If duplicates need to be removed from the first element in the list then:

nList = [[2,5,99,99],[-3,8,1,2,10],[1, 7,100,10]]

output = []

t = []

for e in nList[0]:
    if e not in t:
        t.append(e)
output.append(t)
for e in nList[1:]:
    t = []
    for x in e:
        if x not in output[-1]:
            t.append(x)
    output.append(t)

print(output)

Output:

[[2, 5, 99], [-3, 8, 1, 10], [7, 100]]

Note:

The temptation to use sets should be avoided as the order of the output list may not be as expected

CodePudding user response：

Using this solution you don't even need to make an auxiliary list. It deletes the duplicates in place while the function iterates through the numbers.

This is a much more efficient as it saves the time of building another data structure and saves the space necessary for filling up another nested list for the solution.

nList = [[2,5,99,99],[-3,8,1,2,10],[1, 7,100,10]]

def RemoveDup(nList):
    j = len(nList) - 1          # size of list
    while j >= 0:               # iterate through sublists
        k = len(nList[j]) - 1   # size of sublist
        while k >= 0:           # iterate through each number
            if j - 1 < 0:       # check if there are any lists before position

                # if not then check the current list for duplicates
                if nList[j][k] in nList[j][:k]:  
                    del nList[j][k]  # if there is a duplicate delete it
            
            # if there is a list before the current position check if 
            # it contains the current number
            elif nList[j][k] in nList[j-1]:   
                del nList[j][k]   # if it does delete it
            k -= 1
        j -= 1

RemoveDup(nList)
print(nList)

OUTPUT

[[2, 5, 99], [-3, 8, 1, 10], [7, 100]]

CodePudding user response：

Code:

def RemoveDup(nList):
    tmp =  list(chain.from_iterable(nList))             #Converting nested list to list
    dup = list(set([x for x in tmp if tmp.count(x) > 1]))   #finding the duplicates

    for i in reversed(nList):        #we have reverse the loop so 
          for j in i:                # we can keep first and remove second duplicate
            if j in dup:
                dup.remove(j)        #emptying the dup list as 
                i.remove(j)          # we are deleting the duplicate values
    return nList


RemoveDup(nList)

Output:

[[2, 5, 99], [-3, 8, 1, 10], [7, 100]]

CodePudding user response：

You need to add two intermediate containers

nList = [[2, 5, 99, 99], [-3, 8, 1, 2, 10], [1, 7, 100, 10]]

def RemoveDup(nList):
    aggregation = []
    vessel2 = []
    for i in nList:
        lst = []
        for j in i:
            if j not in aggregation:
                lst.append(j)
                aggregation.append(j)
        vessel2.append(lst)
    return vessel2

print(RemoveDup(nList))