Python for loop not iterating for i=1,2 only outputting first iteration

Question

I am building a for loop using Euler's method for differential equations. The for loop however is not incrementing and only displaying the values for i=0 and not i=1 or i=2.

I have tried manually assigning all arguments and reconstructed the for loop.

import math

def Euler(a,b,N,alpha):
    h=(b-a)/N
    t=a
    w=alpha

    for i in range (0,N):
        w=w+h*(math.exp(t-w))
        t=a+(i*h)
        return t,w    
Euler(0,1,2,1)

I expect the function to return results for i=1 and i=2

your return is inside the for loop. (did you meant to put print there instead?) — pault, Commented Apr 10, 2019 at 18:26
return exits the function immediately, so your loop only executes once. — John Gordon, Commented Apr 10, 2019 at 18:27

b_c · Accepted Answer · 2019-04-10 18:31:47Z

1

As pault mentioned in the comments, your return is inside the loop, which means the function exits on the first iteration.

What you probably want is yield, which would turn the function into a generator:

import math

def Euler(a,b,N,alpha):
    h=(b-a)/N
    t=a
    w=alpha

    for i in range (0,N):
        w=w+h*(math.exp(t-w))
        t=a+(i*h)
        yield t,w

for x, y in Euler(0,1,2,1):
    print(x, y)

>>> 0.0 1.1839397205857212
>>> 0.5 1.3369749844848988

answered Apr 10, 2019 at 18:31

b_c

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What exactly does x,y represent in the second for loop?
– BickenBackBeingBoolean
Commented Apr 11, 2019 at 3:31
@BickenBackBeingBoolean The Euler function/generator yields a tuple (yield t, w). Doing the for x, y in ... unpacks the tuple into two loop variables, instead of having to index a single loop variable (like x[0] or x[1]).
– b_c
Commented Apr 12, 2019 at 13:23

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Eric · Accepted Answer · 2019-04-10 18:30:43Z

0

Your return is in the for loop. Unindent return once.

answered Apr 10, 2019 at 18:30

Eric

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