CODEWITHSUNDEEP
pythonlistloopsif-statement
Lance Collins
Lance Collins

Reputation: 3445

iterating through a list with an if statement

I have a list that I am looping through with a "for" loop and am running each value in the list through an if statement. My problem is that I am trying to only have the program do something if all the values in the list pass the if statement and if one doesn't pass, I want it to move along to the next value in the list. Currently it is returning a value if a single item in the list passes the if statement. Any ideas to get me pointed in the right direction?

Upvotes: 12

Views: 92739

Answers (4)

Rusty Rob
Rusty Rob

Reputation: 17173

You must always be careful if you're deleting items from your list while you're trying to iterate through it.

If you're not deleting then does this help:

>>> yourlist=list("abcdefg")
>>> value_position_pairs=zip(yourlist,range(len(yourlist)))
>>> value_position_pairs
[('a', 0), ('b', 1), ('c', 2), ('d', 3), ('e', 4), ('f', 5), ('g', 6)]
>>> filterfunc=lambda x:x[0] in "adg"
>>> value_position_pairs=filter(filterfunc,value_position_pairs)
>>> value_position_pairs
[('a', 0), ('d', 3), ('g', 6)]
>>> yourlist[6]
'g'

now if value_position_pairs is empty you're done. If not you can increase i by one to go to the next value or iterate through the failed values using their position in the array.

Upvotes: 0

Cédric Julien
Cédric Julien

Reputation: 80761

Maybe you could try with an for ... else statement.

for item in my_list:
   if not my_condition(item):
      break    # one item didn't complete the condition, get out of this loop
else:
   # here we are if all items respect the condition
   do_the_stuff(my_list)

Upvotes: 5

Rob Neuhaus
Rob Neuhaus

Reputation: 9290

You need to loop through your whole list and check the condition before trying to do anything else with the data, so you need two loops (or use some built in that does the loop for you, like all()). From this codepad with nothing too fancy, http://codepad.org/pKfT4Gdc

def my_condition(v):
  return v % 2 == 0

def do_if_pass(l):
  list_okay = True
  for v in l:
    if not my_condition(v):
      list_okay = False

  if list_okay:
    print 'everything in list is okay, including',
    for v in l:
      print v,
    print
  else:
    print 'not okay'

do_if_pass([1,2,3])
do_if_pass([2,4,6])

Upvotes: 0

Martijn Pieters
Martijn Pieters

Reputation: 1121952

Python gives you loads of options to deal with such a situation. If you have example code we could narrow that down for you.

One option you could look at is the all operator:

>>> all([1,2,3,4])
True
>>> all([1,2,3,False])
False

You could also check for the length of the filtered list:

>>> input = [1,2,3,4]
>>> tested = [i for i in input if i > 2]
>>> len(tested) == len(input)
False

If you are using a for construct you can exit the loop early if you come across negative test:

>>> def test(input):
...     for i in input:
...         if not i > 2:
...             return False
...         do_something_with_i(i)
...     return True

The test function above will return False on the first value that's 2 or lower, for example, while it'll return True only if all values were larger than 2.

Upvotes: 13

Related Questions