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How do I optimize nested if statements? I cannot use && here [closed]

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Is there a better way in which this can be written? I'm new to python coding and I want to improve below lines of code. I have included a few lines of my code here but the original code is much longer than this(multiple if statements). Any guidance/feedback would be helpful.
upperchar = #generate a char
pwdlist.append(upperchar)
if len(pwdlist) < length:
upperchar2 = #generate a char
pwdlist.append(upperchar2)
if len(pwdlist) < length:
lowerchar = #generate a char
pwdlist.append(lowerchar)
if len(pwdlist) < length:
lowerchar2 = #generate a char
pwdlist.append(lowerchar2)
if len(pwdlist) < length:
char3 = #generate a char
pwdlist.append(char3)
Posting my first question here so if you need anything more please ask me.

Assuming you are always generating a character the same way, then this would work
pwdlist = []
while len(pwdlist) < length:
pwdlist.append(generate_char())
Or
pwdlist = [generate_char() for _ in range(length)]
Then you can add logic into this generate function to return something based on random values, for example
def generate_char():
from random import randrange
x = randrange(4)
if x == 0:
return 'value a-z'
elif x == 1:
return 'value A-Z'
elif x == 2:
return 'value 0-9'
elif x == 3:
return 'punctuation'

I'm returning a defaultdict(list), but randomly choosing between the two, why does it return nothing sometimes? [closed]

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output = ""
numberList = [0, 1]
print(random.choice(numberList))
if(random.choice(numberList) == 0):
if len(slots) > 0:
output = templates[state][0].replace("<num_classes>", str(slots[0][1]))
else:
output = templates[state][0]
elif(random.choice(numberList) == 1):
if len(slots) > 0:
output = templates2[state][0].replace("<num_classes>", str(slots[0][1]))
else:
output = templates2[state][0]
return output
expected answer
sometimes get this
Sometimes it returns nothing or no dictionary... Why?

With
elif(random.choice(numberList) == 1)
you will again choose a brand new random number. And if that isn't 1 then there's no else that will set output.
Instead of elif you should have a plain else.

Deleting max value in set without max() [closed]

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I have a working code
task = set()
x = 1
while x != 0:
x = int(input('input your number '))
task.add(x)
print('Just 0 could stop it!')
task.remove(max(task))
print(max(task))
And need to get the same result without using max(). What could be an alternative?

Something like this, unless you have really large sets, I don`t see the advantage
task = set()
x = 1
m = 0
while x != 0:
x = int(input('input your number '))
task.add(x)
if x > m:
m = x
print('Just 0 could stop it!')
task.remove(m)
print(max(task))
Notice this will only work for positive numbers, if you want to the complete int range you should init m like this m = -sys.maxsize - 1

You could use min with a key arg that inverts the element:
>>> task = {1, 2, 3, 4, 5}
>>> max(task)
5
>>> min(task, key=lambda x: -x)
5
Or you could sort it and take the last element...
>>> sorted(task)[-1]
5

Compare values stored in variables, then print the variable name [closed]

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I have tried to use the max code but it printed out the largest number not the variable name.
a=3
b=4
c=7
d=6
e=8
max(a,b,c,d,e)
this is the code I've tried but it printed 8 instead of e.
forgot to add that my values for a b c etc came from a loop so im not suppose to know their values myself...

You need to use a dictionary or a named tuple to do something like that.
>>> my_list = {'a': 1, 'b': 2, 'c': 3}
>>> max(my_list, key=my_list.get)
'c'
Let me know if there are any other methods.

Try this
dct={'a':3 ,'b':4 ,'c':7 ,'d':6 ,'e':8}
for i, j in dct.iteritems():
if j == max(dct.values()):
print i

From this question, there seems to be a way to retrieve the name of your variable (see caveat below before using):
a = 3
b = 4
c = 7
d = 6
e = 8
m = max(a, b, c, d, e)
for k, v in list(locals().items()):
if v is m:
v_as_str = k
break
v_as_str
output:
'e'
Caveat:
This may work if you can be sure that there are no other (unrelated) variables with the same value as the max of a to e
somewhere in your code, and appearing earlier in the locals.
Thanks to #D.Everhard & #asongtoruin in the comments

Python: how to display all possible cases to place brackets [closed]

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Given a number of brackets pairs. I want to display all correct combinations of these brackets. By correct I mean each bracket should be opened before closing in each combination. For example, if the number of brackets is 2 the output should be:
(())
()()
For 3:
((()))
()()()
(()())
(())()
()(())
The order of the output lines doesn't matter.
How can I do it with Python?

Try this code, please:
def __F(l, r, pref):
if r < l or l < 0 or r < 0:
return
if r == 0 and l == 0:
print(pref)
return
__F(l - 1, r, pref + "(")
__F(l, r - 1, pref + ")")
def F(n):
__F(n, n, "")
F(2)
F(3)