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Solución:
La solución fácil es:
>>> string1 = 'AJYF'
>>> string2 = 'AJ'
>>> if string2 in string1:
... string1.replace(string2,'')
'YF'
>>>
Creo que lo que quieres es esto:
a = 'AJYF'
b = a.replace('AJ', '')
print a # produces 'YF'
a = 'GTYF'
b = a.replace('GTY', '')
print a # produces 'F'
replace puede hacer algo que no quiere si el segundo string está presente en varias posiciones:
s1 = 'AJYFAJYF'
s2 = 'AJ'
if s1.startswith(s2):
s3 = s1.replace(s2, '')
s3
# 'YFYF'
Puede agregar un argumento adicional a replace para indicar que desea que solo ocurra un reemplazo:
if s1.startswith(s2):
s3 = s1.replace(s2, '', 1)
s3
# 'YFAJYF'
O podrías usar el re módulo:
import re
if s1.startswith(s2):
s3 = re.sub('^' + s2, '', s1)
s3
# 'YFAJYF'
El '^' es asegurar que s2 se sustituye sólo en la primera posición de s1.
Otro enfoque más, sugerido en los comentarios, sería eliminar el primer len(s2) personajes de s1:
if s1.startswith(s2):
s3 = s1[len(s2):]
s3
# 'YFAJYF'
Algunas pruebas que utilizan la magia %timeit en ipython (python 2.7.12, ipython 5.1.0) sugieren que este último enfoque es más rápido:
In [1]: s1 = 'AJYFAJYF'
In [2]: s2 = 'AJ'
In [3]: %timeit s3 = s1[len(s2):]
The slowest run took 24.47 times longer than the fastest. This could mean that an intermediate result is being cached.
10000000 loops, best of 3: 87.7 ns per loop
In [4]: %timeit s3 = s1[len(s2):]
The slowest run took 32.58 times longer than the fastest. This could mean that an intermediate result is being cached.
10000000 loops, best of 3: 87.8 ns per loop
In [5]: %timeit s3 = s1[len(s2):]
The slowest run took 21.81 times longer than the fastest. This could mean that an intermediate result is being cached.
10000000 loops, best of 3: 87.4 ns per loop
In [6]: %timeit s3 = s1.replace(s2, '', 1)
The slowest run took 17.64 times longer than the fastest. This could mean that an intermediate result is being cached.
1000000 loops, best of 3: 230 ns per loop
In [7]: %timeit s3 = s1.replace(s2, '', 1)
The slowest run took 17.79 times longer than the fastest. This could mean that an intermediate result is being cached.
1000000 loops, best of 3: 228 ns per loop
In [8]: %timeit s3 = s1.replace(s2, '', 1)
The slowest run took 16.27 times longer than the fastest. This could mean that an intermediate result is being cached.
1000000 loops, best of 3: 234 ns per loop
In [9]: import re
In [10]: %timeit s3 = re.sub('^' + s2, '', s1)
The slowest run took 82.02 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 1.85 µs per loop
In [11]: %timeit s3 = re.sub('^' + s2, '', s1)
The slowest run took 12.82 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 1.86 µs per loop
In [12]: %timeit s3 = re.sub('^' + s2, '', s1)
The slowest run took 13.08 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 1.84 µs per loop
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