HI, I am a real beginner, so please, be gentle:)
I wrote a short programm:
bits_string = "01010102"
bits = ["0", "1"]
bits_list = []
for i in bits_string:
bits_list.append(i)
result = all(i in bits_list for i in bits)
print(result)And the outcome is:
Output:
['0', '1', '0', '1', '0', '1', '0', '2']
True
I mean, why is it "TRUE"? "2" is not present in list "bits".
Thanks in advance
because both '0' and '1' are present in bits_list
Your expected output is consistent with all(i in bits for i in bits_list) - this will be False, because of 2 not present in bits.
That said there is no need to convert the string to list you can iterate over str just fine.
Your logic is backward. If you want verify that all the characters from bits_string are in bits, the test should be:
result = all(i in bits for i in bits_list)
As buran mentioned there is no need to convert bits_string to bits_list, or to have a list at all. Iterators and "in" for str work very similar to how they work for list. You could write the code like this:
def valid_binary_string(binary_string):
return all(bit in "01" for bit in binary_string)
print(valid_binary_string("01010100"))
print(valid_binary_string("01010102"))There are other ways to solve this problem that may be more efficient. For example, you could use sets to reduce the number of comparisons.
def valid_binary_string(binary_string):
return all(i in '01' for i in set(binary_string))
print(valid_binary_string("01010100"))
print(valid_binary_string("01010102"))If you use sets you can use set operations. For example, set(binary_string) - set('01') will return an empty set for any vailid binary string.
binary_digits = set('01')
def valid_binary_string(binary_string):
return not set(binary_string)-binary_digits
print(valid_binary_string("01010100"))
print(valid_binary_string("01010102"))