Question about lambdas in python

Question

The following code produces 222. Why?
def bla():
return [lambda x:i*x for i in range(3)]
for b in bla():
print(b(1), end='')

I thought it should produce 012, as bla() returns a list of three functions that do 0*x, 1*x and 2*x respectively. Then in the for-loop we pass 1 to them, and should get 012... But we get 222, why?

Answer 1

@khao It exactly does. But by writing for b in bla() you kind of iterate through them. And remember that bla() itself does not take arguments, just returns the list of anonymous functions. Each of those functions take 1 as argument, because of b(1). And since those lambdas were not told to return anything, they just assume the last generated value, each.

And this is what you print.

Answer 2

Only the last item in lambda is returned. It equals 2 and is printed 3 times in a row.
If you used b(5) it should print 101010

Answer 3

@merkrafter Yes, exactly. And then it is iterated through, but does not specifically return anything. When done executing it just assumes the last value and this is the one that is printed.

Answer 4

I think its because x is string,so it will concat it 3 times

Answer 5

@khao, What would you like it to return?

bla() itself does not take any argument, but is of a function class. If you call one of its instances (for b in bla()) AND you provide an argument (1), it goes to that function and carries out its iteration. When it's done, it returns the last value it reached, becuase you told it to return a value, not the function itself.

Answer 6

@Kuba, I thought my bla() would contain three different lambda functions, and when I pass an argument in a for loop, it would loop through these three funcs and produce respective outputs... Your explanation makes sense, but then I don't understand why each one instance of bla() doesn't produce a list of three lambdas, but an integer?

Answer 7

@Kuba ok, now I get it. The crucial missing bit for me was that the lambdas would assume the last value because not explicitly returning anything. Thanks a lot!

Answer 8

how is x string? isn't x = 1 here?

Answer 9

I don't know the answer (yet), but if you vary that n in range(n) (in your case it's n=3) you'll get
str(n-1)*n as your output.
So for n=4 the output is 3333

AND: print (type(b(1))) proofes that you get integers

Answer 10

@merkrafter yeah, it looks as if the n lambda funcs in the list returned by bla() are all the same, namely taking the last i in range, which is n-1. The question is why...

Answer 11

@Kuba but why is only the last lambda returned?

Answer 12

So bla() does NOT return
[lambda x: 0*x, lambda x: 1*x, lambda x: 2*x] but
[lambda x: i*x for... and that three times, right?