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How to remove elements from a list that appear less than k = 2?

I am trying to keep elements of a list that appear at least twice, and remove the elements that appear less than twice.
For example, my list can look like:
letters = ['a', 'a', 'b', 'b', 'b', 'c']
I want to get a list with the numbers that appear at least twice, to get the following list:
letters_appear_twice = ['a', 'b'].
But since this is part of a bigger code, I don't know exactly what my lists looks like, only that I want to keep the letters that are repeated at least twice. But for the sake of understanding, we can assume I know what the list looks like!
I have tried the following:
'''
letters = ['a', 'a', 'b', 'b', 'b', 'c']
for x in set(letters):
if letters.count(x) > 2:
while x in letters:
letters.remove(x)
print(letters)
'''
But this doesn't quite work like I want it too...
Thank you in advance for any help!

letters = ['a', 'a', 'b', 'b', 'b', 'c']
res = []
for x in set(letters):
if letters.count(x) >= 2:
res.append(x)
print(res)
Prints:
['b', 'a']

Using your code above. You can make a new list, and append to it.
new_list = []
for x in set(letters):
if letters.count(x) >= 2:
new_list.append(x)
print(new_list)
Output
['b', 'a']

Easier to create a new list instead of manipulating the source list
def letters_more_or_equal_to_k(letters, k):
result = []
for x in set(letters):
if letters.count(x) >= k:
result.append(x)
result.sort()
return result
def main():
letters = ['a', 'a', 'b', 'b', 'b', 'c']
k = 2
result = letters_more_or_equal_to_k(letters, k)
print(result) # prints ['a', 'b']
if __name__ == "__main__":
main()

If you don't mind shuffling the values, here's one possible solution:
from collections import Counter
letters = ['a', 'a', 'b', 'b', 'b', 'c']
c = Counter(letters)
to_remove = {x for x, i in c.items() if i < 2}
result = list(set(letters) - to_remove)
print(result)
Output:
['a', 'b']
You can always sort later.
This solution is efficient for lists with more than ~10 unique elements.

Combining values in an array in Python

I have an array like so: ['A', 'B', 'C', 'D', 'E'].
I've been trying to figure out how to make it like so: ['AB', 'CD', 'E']
I'm not sure where to start. Thanks in advance!

main.py
a = ['A', 'B', 'C', 'D', 'E']
b = [i + j for i, j in zip(a[:-1:2], a[1::2])]
if len(a) % 2 == 1:
b.append(a[-1])
print(b)
result
$ python main.py
['AB', 'CD', 'E']

>>> [''.join(a[i:i+2]) for i in range(0, len(a), 2)]
['AB', 'CD', 'E']
or (as I love iterators)
>>> it = iter(a)
>>> [s + next(it, '') for s in it]
['AB', 'CD', 'E']

I think that the easier way is to iterate through the array and concatenate the chars, it works if you have an array with even length, so you could add a check and append the last element in case of odd length.
array = ['A', 'B', 'C', 'D', 'E']
array2 = [f"{array[i]}{array[i+1]}" for i in range(0, len(array)-1, 2)]
if len(array)%2!=0:
array2.append(array[-1])
print(array2)

Try like this. This is very bare answer but should work.
my_array = ['A', 'B', 'C', 'D', 'E']
def combine_array(my_array):
mixed_array = []
start_new = True
for item in my_array:
if start_new:
mixed_array.append(item)
start_new = False
else:
mixed_array[-1] = mixed_array[-1] + item
start_new = True
return mixed_array
if __name__ == "__main__":
try:
print(combine_array(my_array))
except Exception as err:
print(err)

startArray = ['A', 'B', 'C', 'D', 'E']
currentIndex = 0
finishArray = ['']
for x in startArray:
if len(finishArray[currentIndex]) == 2:
currentIndex += 1
finishArray.insert(currentIndex,x)
else:
finishArray[currentIndex] += x
print(finishArray)

Putting column values from text file into a list in python

I have a text file like this:
a w
b x
c,d y
e,f z
And I want to get the values of the first column into a list without duplicates. For now I get the values from the first column, which I am doing like this:
f=open("file.txt","r")
lines=f.readlines()
firstCol=[]
for x in lines:
firstCol.append(x.split(' ')[0])
f.close()
In the next step I want to separate the values by a comma delimiter the same way I did before, but then I get an output like this:
[['a'], ['b'], ['c', 'd'], ['e', 'f']]
How can I convert this into a one dimensional thing to be able to remove duplicates afterwards?
I am a beginner in python.

You can split it immediately after the first split and must use extend instead of append.
f=open("file.txt","r")
lines=f.readlines()
firstCol=[]
for x in lines:
firstCol.extend(x.split(' ')[0].split(','))
f.close()
print(firstCol)
Result
['a', 'b', 'c', 'd', 'e', 'f']
Or if you want to keep the firstCol
f=open("file.txt","r")
lines=f.readlines()
firstCol=[]
for x in lines:
firstCol.append(x.split(' ')[0])
f.close()
one_dimension = []
for col in firstCol:
one_dimension.extend(col.split(','))
print(firstCol)
print(one_dimension)
Result
['a', 'b', 'c,d', 'e,f']
['a', 'b', 'c', 'd', 'e', 'f']

you can use itertools.chain to flatten your list of lists and then you can use the built-in class set to remove the duplicates :
from itertools import chain
l = [['a'], ['b'], ['c', 'd'], ['e', 'f']]
set(chain.from_iterable(l))
# {'a', 'b', 'c', 'd', 'e', 'f'}
to flatten your list you can also use a list comprehension:
my_l = [e for i in l for e in i]
# ['a', 'b', 'c', 'd', 'e', 'f']
same with 2 simple for loops:
my_l = []
for i in l:
for e in i:
my_l.append(e)

Possible solution 1
If your are fine with your code, you can keep like that and remove duplicates from a list of lists executing the following:
import itertools
firstCol.sort()
firstCol = list(x for x,_ in itertools.groupby(firstCol))
Possible solution 2
If you want to convert the list of lists into one list of items:
firstCol = [x for y in firstCol for x in y]
If you want to also remove duplicates:
firstCol = list(set([x for y in firstCol for x in y]))

replace duplicate values in a list with 'x'?

I am trying to understand the process of creating a function that can replace duplicate strings in a list of strings. for example, I want to convert this list
mylist = ['a', 'b', 'b', 'a', 'c', 'a']
to this
mylist = ['a', 'b', 'x', 'x', 'c', 'x']
initially, I know I need create my function and iterate through the list
def replace(foo):
newlist= []
for i in foo:
if foo[i] == foo[i+1]:
foo[i].replace('x')
return foo
However, I know there are two problems with this. the first is that I get an error stating
list indices must be integers or slices, not str
so I believe I should instead be operating on the range of this list, but I'm not sure how to implement it. The other being that this would only help me if the duplicate letter comes directly after my iteration (i).
Unfortunately, that's as far as my understanding of the problem reaches. If anyone can provide some clarification on this procedure for me, I would be very grateful.

Go through the list, and keep track of what you've seen in a set. Replace things you've seen before in the list with 'x':
mylist = ['a', 'b', 'b', 'a', 'c', 'a']
seen = set()
for i, e in enumerate(mylist):
if e in seen:
mylist[i] = 'x'
else:
seen.add(e)
print(mylist)
# ['a', 'b', 'x', 'x', 'c', 'x']

Simple Solution.
my_list = ['a', 'b', 'b', 'a', 'c', 'a']
new_list = []
for i in range(len(my_list)):
if my_list[i] in new_list:
new_list.append('x')
else:
new_list.append(my_list[i])
print(my_list)
print(new_list)
# output
#['a', 'b', 'b', 'a', 'c', 'a']
#['a', 'b', 'x', 'x', 'c', 'x']

The other solutions use indexing, which isn't necessarily required.
Really simply, you could check if the value is in the new list, else you can append x. If you wanted to use a function:
old = ['a', 'b', 'b', 'a', 'c']
def replace_dupes_with_x(l):
tmp = list()
for char in l:
if char in tmp:
tmp.append('x')
else:
tmp.append(char)
return tmp
new = replace_dupes_with_x(old)

You can use the following solution:
from collections import defaultdict
mylist = ['a', 'b', 'b', 'a', 'c', 'a']
ret, appear = [], defaultdict(int)
for c in mylist:
appear[c] += 1
ret.append(c if appear[c] == 1 else 'x')
Which will give you:
['a', 'b', 'x', 'x', 'c', 'x']

How can I reverse a section of a list using a loop in Python?

I'm in need of a little help reversing a section of a list in Python using a loop.
I have a list: mylist = ['a', 'b', 'c', 'd', 'e', 'f']
Also have an index number, this number will tell where to start the reversing. For example, if the reverse-index number is 3, it needs to be something like this: ['d', 'c', 'b', 'a', 'e', 'f']
What I currently have:
def list_section_reverse(list1, reverse_index):
print("In function reverse()")
n_list = []
for item in range( len(list1) ):
n_list.append( (list1[(reverse_index - 1) - item]) )
item += 1
return n_list
mylist = ['a', 'b', 'c', 'd', 'e', 'f']
print( list_section_reverse(mylist, 3) )
Which returns ['c', 'b', 'a', 'f', 'e', 'd']
How can I alter my code, so that it prints out ['d', 'c', 'b', 'a', 'e', 'f']?

You can simply use:
def list_section_reverse(list1, reverse_index):
return list(reversed(list1[:reverse_index+1])) + list1[reverse_index+1:]
Edit: The problem with your existing solution is that you keep reversing after reverse_index. If you have to use a loop, try this:
def list_section_reverse(list1, reverse_index):
print("In function reverse()")
n_list = list1[:]
for i in range(reverse_index + 1):
n_list[i] = list1[-i-reverse_index]
return n_list
mylist = ['a', 'b', 'c', 'd', 'e', 'f']
print(list_section_reverse(mylist, 3))

The pythonic solution:
list1[reverse_index::-1] + list1[reverse_index+1:]
Now, that's not using loops like you asked. Well, not explicitly... Instead we can break down the above into its constituent for loops.
def list_section_reverse(list1, reverse_index):
if reverse_index < 0 or reversed_index >= len(list1):
raise ValueError("reverse index out of range")
reversed_part = []
for i in range(reverse_index, -1, -1): # like for i in [n, n-1, ..., 1, 0]
reversed_part.append(list1[i]
normal_part = []
for i in range(reverse_index + 1, len(list1)):
normal_part.append(list1[i])
return reversed_part + normal_part

Is it allowed to make a copy of the list?
def list_section_reverse(list1, reverse_index):
print("In function reverse()")
n_list = [ element for element in list1 ]
for item in range( reverse_index + 1 ):
n_list[ item ] = list1[ reverse_index - item ]
item += 1
return n_list
mylist = ['a', 'b', 'c', 'd', 'e', 'f']
print(list_section_reverse(mylist, 3))
Outs:
In function reverse()
['d', 'c', 'b', 'a', 'e', 'f']

You can modify the list inplace using a slice.
mylist[:4] = mylist[:4][::-1]

You can try this. This makes use of no slicing, and can use either a while loop or for loop.
def reversed_list(my_list, index):
result = []
list_copy = my_list.copy()
i = 0
while i < index+1:
result.append(my_list[i])
list_copy.remove(my_list[i])
i+=1
result.reverse()
return result + list_copy
Or with a for loop
def reversed_list(my_list, index):
result = []
list_copy = my_list.copy()
for i in range(len(my_list)):
if i < index + 1:
result.append(my_list[i])
list_copy.remove(my_list[i])
result.reverse()
return result + list_copy