Return True if the given string contains an appearance of "xyz" where the xyz is not directly preceeded by a period(.). So "xxyz" is True but "x.xyz" is False.

Question

One of the many correct answers is:

def xyz_there(str):
for i in range(len(str)):
if str[i] != '.' and str[i+1:i+4] == 'xyz':
return True
if str[0:3] == "xyz":
return True
return False

Can anyone explain what the line of code means on line 3? Thanks in advance.

Source of qn: codingbat.com

Answer 1

str[i] != '.' checks if the character at index i is not a dot.
str[i+1:i+4] == 'xyz' checks if the substring from indexes i+1 to i+3 is "xyz".

Answer 2

@Kelvin: Yes, != is the operator to check if the two operands are not equal.

Answer 3

Zen is right. the third line you are asking is that very logic you need to implement to get the result.

Answer 4

So is it safe to say that "!=" is exactly "not =="?
etc. str[i] == '.' means that both of these operands are equivalent

Answer 5

Ok it's all clear now. Thank you very much =)

Answer 6

and you dont need to iterate i up to len(str) but up to len(str)-3

Answer 7

hi