CODEWITHSUNDEEP
python
Pablo
Pablo

Reputation: 5087

Python, next iteration of the loop over a loop

I need to get the next item of the first loop given certain condition, but the condition is in the inner loop. Is there a shorter way to do it than this? (test code)

    ok = 0
    for x in range(0,10):
        if ok == 1:
            ok = 0
            continue
        for y in range(0,20): 
            if y == 5:
                ok = 1
                continue

What about in this situation?

for attribute in object1.__dict__:
    object2 = self.getobject()
    if object2 is not None:
        for attribute2 in object2: 
            if attribute1 == attribute2:
                # Do something
                #Need the next item of the outer loop

The second example shows better my current situation. I dont want to post the original code because it's in spanish. object1 and object2 are 2 very different objects, one is of object-relational mapping nature and the other is a webcontrol. But 2 of their attributes have the same values in certain situations, and I need to jump to the next item of the outer loop.

Upvotes: 15

Views: 43495

Answers (5)

telliott99
telliott99

Reputation: 7907

It's not completely clear to me what you want, but if you're testing something for each item produced in the inner loop, any might be what you want (docs)

>>> def f(n):  return n % 2
... 
>>> for x in range(10):
...     print 'x=', x
...     if f(x):
...         if any([y == 8 for y in range(x+2,10)]):
...             print 'yes'
... 
x= 0
x= 1
yes
x= 2
x= 3
yes
x= 4
x= 5
yes
x= 6
x= 7
x= 8
x= 9

Upvotes: 0

MattH
MattH

Reputation: 38247

So you're not after something like this? I'm assuming that you're looping through the keys to a dictionary, rather than the values, going by your example.

for i in range(len(object1.__dict__)):
  attribute1 = objects1.__dict__.keys()[i]
  object2 = self.getobject() # Huh?
  if object2 is not None:
    for j in range(len(object2.__dict__)):
      if attribute1 == object2.__dict__.keys()[j]:
        try:
          nextattribute1 = object1.__dict__.keys()[i+1]
        except IndexError:
          print "Ran out of attributes in object1"
          raise

Upvotes: 0

jfs
jfs

Reputation: 414139

Your example code is equivalent to (that doesn't seem what you want):

for x in range(0, 10, 2):
    for y in range(20): 
        if y == 5:
           continue

To skip to the next item without using continue in the outer loop:

it = iter(range(10))
for x in it:
    for y in range(20):
        if y == 5:
           nextx = next(it)
           continue

Upvotes: 2

MAK
MAK

Reputation: 26586

Replace the continue in the inner loop with a break. What you want to do is to actually break out of the inner loop, so a continue there does the opposite of what you want.

ok = 0
for x in range(0,10):
    print "x=",x
    if ok == 1:
        ok = 0
        continue
    for y in range(0,20): 
        print "y=",y
        if y == 5:
            ok = 1
            break

Upvotes: 16

ron
ron

Reputation: 9354

You can always transform into a while loop:

flag = False
for x in range(0, 10):
    if x == 4: 
        flag = True
        continue

becomes

x = 0
while (x != 4) and x < 10:
    x += 1
flag = x < 10

Not necessary simpler, but nicer imho.

Upvotes: 1

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