Method of Python list to remove duplication and keep the original order unchanged

Directory background
code & & demo
Summarize

Background

python's weight removal operation is as fierce as a tiger, set list pulls down, and it is weight removal, but the order is disrupted. If there is no need for order, it really doesn't matter.
But if you need to keep the order, you need one small change.

code & & demo

list is duplicated and the order is out of order

# normal  Writing style 
l1 = ['b','c','d','b','c','a','a']
l2 = list(set(l1))
print(l2)

# plus  Writing style 
l1 = ['b','c','d','b','c','a','a']
l2 = {}.fromkeys(l1).keys()

After duplication removal, it is still the original list sequence

# normal  Writing style 
l1 = ['b','c','d','b','c','a','a']
l2 = list(set(l1))
l2.sort(key=l1.index)

# plus  Writing style 
l1 = ['b','c','d','b','c','a','a']
l2 = sorted(set(l1),key=l1.index)

Realization of writing loop code

# normal  Writing style 
l1 = ['b','c','d','b','c','a','a']
l2 = []
for i in l1:
  if not i in l2:
    l2.append(i)
    
# plus  Writing style 

l1 = ['b','c','d','b','c','a','a']
l2 = []
[l2.append(i) for i in l1 if not i in l2]

Implementation of writing while loop code

L = [3, 1, 2, 1, 3, 4]
T = L[:]
for i in L:
  while T.count(i) > 1:
    del T[T.index(i)]
T.sort(key=L.index)

lambda writing

Remarks:

ambda L, i: L if i in L L + [i] # Returns the list itself if the element is in the list, L + [i] if it is not [[],] + L # is equivalent to [[], L], which is convenient for later calculation

Summarize

If you struggle with spatial complexity, what do you do with python?
Let's talk about whether it can be completed first, and then talk about optimization.

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