In Python 3.5, given this string:
"rsFooBargrdshtrshFooBargreshyershytreBarFootrhj"
and the index 17 -- so, the F at the start of the second occurrence of FooBar -- how can I check that "FooBar" exists? In this case, it should return True, while if I gave it the index 13 it should return false.
There's actually a very simple way to do this without using any additional memory:
>>> s = "rsFooBargrdshtrshFooBargreshyershytreBarFootrhj"
>>> s.startswith("FooBar", 17)
True
>>>
The optional second argument to startswith tells it to start the check at offset 17 (rather than the default 0). In this example, a value of 2 will also return True, and all other values will return False.
You need to slice your original string based on your substring's length and compare both the values. For example:
>>> my_str = "rsFooBargrdshtrshFooBargreshyershytreBarFootrhj"
>>> word_to_check, index_at = "FooBar", 17
>>> word_to_check == my_str[index_at:len(word_to_check)+index_at]
True
>>> word_to_check, index_at = "FooBar", 13
>>> word_to_check == my_str[index_at:len(word_to_check)+index_at]
False
print("rsFooBargrdshtrshFooBargreshyershytreBarFootrhj"[17:].startswith('Foo')) # True
or in common
my_string[start_index:].startswith(string_to_check)
Using Tom Karzes approach, as a function
def contains_at(in_str, idx, word):
return in_str[idx:idx+len(word)] == word
>>> contains_at(s, 17, "FooBar")
>>> True
Try this:
def checkIfPresent(strng1, strng2, index):
a = len(strng2)
a = a + index
b = 0
for i in range(index, a):
if strng2[b] != strng1[i]:
return false
b = b+1
return true
s = "rsFooBargrdshtrshFooBargreshyershytreBarFootrhj"
check = checkIfPresent(s, Foobar, 17)
print(check)