I want to do something like this:
>>> mystring = \"foo\"
>>> print(mid(mystring))
Help!
This is Andy's solution. I just addressed User2357112's concern and gave it meaningful variable names. I'm a Python rookie and preferred these functions.
def left(aString, howMany):
if howMany <1:
return ''
else:
return aString[:howMany]
def right(aString, howMany):
if howMany <1:
return ''
else:
return aString[-howMany:]
def mid(aString, startChar, howMany):
if howMany < 1:
return ''
else:
return aString[startChar:startChar+howMany]
slices to the rescue :)
def left(s, amount):
return s[:amount]
def right(s, amount):
return s[-amount:]
def mid(s, offset, amount):
return s[offset:offset+amount]
There are built-in functions in Python for "right" and "left", if you are looking for a boolean result.
str = "this_is_a_test"
left = str.startswith("this")
print(left)
> True
right = str.endswith("test")
print(right)
> True
You can use this method also it will act like that
thadari=[1,2,3,4,5,6]
#Front Two(Left)
print(thadari[:2])
[1,2]
#Last Two(Right)# edited
print(thadari[-2:])
[5,6]
#mid
mid = len(thadari) //2
lefthalf = thadari[:mid]
[1,2,3]
righthalf = thadari[mid:]
[4,5,6]
Hope it will help
Thanks Andy W
I found that the mid() did not quite work as I expected and I modified as follows:
def mid(s, offset, amount):
return s[offset-1:offset+amount-1]
I performed the following test:
print('[1]23', mid('123', 1, 1))
print('1[2]3', mid('123', 2, 1))
print('12[3]', mid('123', 3, 1))
print('[12]3', mid('123', 1, 2))
print('1[23]', mid('123', 2, 2))
Which resulted in:
[1]23 1
1[2]3 2
12[3] 3
[12]3 12
1[23] 23
Which was what I was expecting. The original mid() code produces this:
[1]23 2
1[2]3 3
12[3]
[12]3 23
1[23] 3
But the left() and right() functions work fine. Thank you.
If I remember my QBasic, right, left and mid do something like this:
>>> s = '123456789'
>>> s[-2:]
'89'
>>> s[:2]
'12'
>>> s[4:6]
'56'
http://www.angelfire.com/scifi/nightcode/prglang/qbasic/function/strings/left_right.html