How to check if a list has both empty and non-empty values in python

Question

I am trying to check if a list has both empty and non-empty values. My code is as follows.

list = ['','a','b','']
not_empth = not float('NaN')
if float('NaN') and not_empth in list:
 print('List has both empty and non-empty values.')
else:
 print('List does not have both empty and non-empty values.')

However, the output shows as below, which is not correct.

List does not have both empty and non-empty values.

• j1-lee: What is float('NaN') doing here? Your list only has strings.

Answer 1

This checks to see if a list contains both empty strings and non-empty strings

list = ['','a','b','']
has_empty = any(s == '' for s in list) #for items in list, if any equal '' then return true
has_non_empty = any(s != '' for s in list) #for items in list, if any do not equal '' then return true
if has_empty and has_non_empty:
 print('List has both empty and non-empty values.')
else:
 print('List does not have both empty and non-empty values.')

Answer 2

list = ["Efsf","efsfs",""]
list1 = ["",'']
list2 = ["fsef",5 ,5]
def check(list):
 null =''
 if null in list and len(list)-list.count(null)>0:
 print("List has both empty and non-empty values.")
 else:
 print("List does not have both empty and non-empty values.")
check(list)
check(list1)
check(list2)

output:

List has both empty and non-empty values.

List does not have both empty and non-empty values.

List does not have both empty and non-empty values.

Answer 3

You could use a for loop, as a different approach to this. Please note I changed float("NaN") to '' due to the reason @j1-lee said.

PLEASE NOTE: This approach is not recommended if you have a large list.

list = ['','a','b','']
check_1 = 0 # 0 being represented as false.
check_2 = 0 
for item in list:
 if item == '':
 check_1 = 1
 else:
 check_2 = 1
if check_1 and check_2 == 1:
 print(True)
else:
 print(False)

Output: True

• Chris: No need to use 0 or 1 to represent boolean values, as Python has True and False.
• Chris: I would suggest that you break your loop if both check_1 and check_2 are True. No further checks are necessary if that condition is met.
• Brody Critchlow: @Chris Yes, this was just a start so he could expand on it.

Answer 4

What you have in the question just needs an if in operation to solve.

>>> list_ = ['','a','b','']
>>> if '' in list_:
... print('List has both empty and non-empty values.')
... elif list_ == []:
... print('List does not have both empty and non-empty values.')
...
List has both empty and non-empty values.

As long you have at least one '', the first condition will be fulfilled. For your second print statement, if I understand correctly, it is to check an empty list.

Answer 5

Having some fun with functools.reduce to accomplish this in one pass over the list.

lst = ['', 'a', 'b', '']
reduce(lambda acc, x: (acc[0], True) if x != '' else (True, acc[1]), 
 lst, (False, False))
# (True, True)
all(reduce(lambda acc, x: (acc[0], True) if x != '' else (True, acc[1]), 
 lst, (False, False)))
# True

Of course, this does have to evaluate every element in the list and cannot short-circuit. To do that, we might implement a function reduce_until which lets us supply a function that provides a condition for early exit.

def reduce_until(f, cond, iter, init):
 for i in iter:
 if cond(init): return init
 init = f(init, i)
 return init
reduce_until(lambda acc, x: (acc[0], True) if x != '' else (True, acc[1]),
 all, lst, (False, False))
# (True, True)
all(reduce_until(lambda acc, x: (acc[0], True) if x != '' else (True, acc[1]),
 all, lst, (False, False))
# True