CODEWITHSUNDEEP
pythoniteratorgenerator
prongs
prongs

Reputation: 9606

Python generator behaviour

import itertools
ws=[]
subs=[]
set_subs=[]
for i in xrange(int(raw_input())):
    S=raw_input()
    l=len(S)
    subs.append(S[i:j+1] for i in xrange(l) for j in xrange(i,l))

input: 

2
aab
aac

now both subs[0] and subs[1] give me same result.

print list(subs[0])
>>>['a','aa','aac','a','ac','c']
print list(subs[1])
>>>['a','aa','aac','a','ac','c']

whereas list(subs[0]) should have been ['a','aa','aab','a','ab','b']

I vaguely understand why this is happening. What do I do to make subs[0] and subs[1] actually different.

NOTE: changing the line

subs.append(S[i:j+1] for i in xrange(l) for j in xrange(i,l))

with 

subs.append([S[i:j+1] for i in xrange(l) for j in xrange(i,l)])

is not an option

Upvotes: 2

Views: 198

Answers (2)

Janne Karila
Janne Karila

Reputation: 25197

http://docs.python.org/reference/expressions.html#generator-expressions

Variables used in the generator expression are evaluated lazily when the __next__() method is called for generator object (in the same fashion as normal generators). However, the leftmost for clause is immediately evaluated, so that an error produced by it can be seen before any other possible error in the code that handles the generator expression. Subsequent for clauses cannot be evaluated immediately since they may depend on the previous for loop.

S[i:j+1] is evaluated when you execute the generator, and at that point S has the latest value.

You can use a normal generator instead. Now ss is local to subgen:

import itertools

def subgen(ss):
    l=len(ss)
    for i in xrange(l):
        for j in xrange(i,l):
            yield ss[i:j+1]

subs=[]
for i in xrange(int(raw_input())):
    S=raw_input()
    subs.append(subgen(S))

Upvotes: 6

Mariy
Mariy

Reputation: 5914

Well, this is one of the strange features of generator expression. Have a look at this In your case it is used late binding and that's why you will get two identical results.

Upvotes: 2

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