Comparing True value in list to another list-CodePudding

lst = [1, 3, 2, 1, 3, True, False, True, "apple", "banana", "mango", "apple"]

lst2 = []
for i in range(0, len(lst)):
    if lst[i] not in lst2:
        lst2.append(lst[i])
print(lst2)

Above code is to remove duplicate values. But it is also removing True value. Anyone Explain me why it's not storing True value and show me correct code.

Desired Output:

[1,3,2, True, False, “apple”, “banana”, “mango”]

Output I'm getting:

[1,3,2, False, “apple”, “banana”, “mango”]

Can anyone help me with this and explain me reason behind it. Please it would be really helpful.

CodePudding user response：

In Python True is considered equivalent to 1 for the purposes of equality checks and mathematical operations, even though they have different types (bool vs int).

If you want to compare them as being different, a workaround might be to convert each value into a tuple with its type (because even though 1 == True, (1, int) != (True, bool)) for purposes of the comparison, and then convert them back when you're done:

>>> lst = [1, 3, 2, 1, 3, True, False, True, "apple", "banana", "mango", "apple"]
>>> for i in lst:
...     if (i, type(i)) not in lst2:
...         lst2.append((i, type(i)))
...
>>> lst2 = [i for i, _ in lst2]
>>> lst2
[1, 3, 2, True, False, 'apple', 'banana', 'mango']

If you don't care about the order, you can do this more easily with a set:

>>> [i for i, _ in {(i, type(i)) for i in lst}]
[True, 2, 3, 'mango', False, 'banana', 1, 'apple']

or if you do care about the order but still want to do it in a comprehension, you can do a bit of trickery with a dict (since dict keys are unique but unlike a set they preserve insertion order):

>>> [i for i, _ in {(i, type(i)): None for i in lst}]
[1, 3, 2, True, False, 'apple', 'banana', 'mango']

CodePudding user response：

You can use this code.


lst = [1, 3, 2, 1, 3, True, False, True, "apple", "banana", "mango", "apple"]

lst2 = []
for a in lst:
    if not any(a is i for i in lst2):
        lst2.append(a)

print(lst2)

OUTPUT [1, 3, 2, True, False, 'apple', 'banana', 'mango']

CodePudding user response：

Like @Samwise said, when python compares booleans it actually sets them to integers: True == 1 and False == 0. You can try to process them separatly like this:

lst = [1, 3, 2, 1, 3, True, False, True, "apple", "banana", "mango", "apple"]

lst_bools = list(set([i for i in lst if type(i) == bool]))
lst_not_bool = list(set([i for i in lst if type(i) != bool]))
lst2 = lst_bools   lst_not_bool

CodePudding user response：

This appears to work and also retain the list order.

lst = [1, 3, 2, 1, 3, True, False, True, "apple", "banana", "mango", "apple"]
s = set((e, type(e)) for e in lst)
newlist = []
for e in lst:
    if (t := (e, type(e))) in s:
        newlist.append(t[0])
        s.remove(t)
print(newlist)

Output:

[1, 3, 2, True, False, 'apple', 'banana', 'mango']