For in loop and if statement in Python

Question

I'm Trying to figure out how this works. Example One:

n = ["1ab", "2an", "3bca", "4adc"]
l = ["1", "2", "3"]
for m in n:
 if "a" in m:
 for k in l:
 if k in m:
 print k
1
2
3

Now I will try to print last member of n list.

n = ["1ab", "2an", "3bca", "4adc"]
l = ["1", "2", "3"]
for m in n:
 if "a" in m:
 for k in l:
 if not k in m:
 print k
2
3
1
3
1
2
1
2
3

I need to print a list member which does not contain any number listed in l variable but contains "a" in it.

• Joucks: Please explain a bit more what you are trying to do
• IKA: I need to print a list member which does not contain any number listed in l variable but contains "a" in it.
• Joucks: Then you should edit your question

Answer 1

Since 4 is not in your list l , you cannot print it.

• IKA: Thanks for reply. So how to print it?
• Srinivas Reddy Thatiparthy: @iRex it helps a bit more if you say what your are trying to achieve?
• IKA: @ Srinivas Reddy Thatiparthy I need to compare if variable n contains any number of indicated in the variable l. Then Print them separately. In second example I need to print ""4adc".

Answer 2

n = ["1ab", "2an", "3bca", "4adc"]
l = ["1", "2", "3"]
for m in n:
 if "a" in m:
 if not any([k in m for k in l]):
 print m
4adc

Answer 3

The answer is in your second to last line. For every loop through n you loop through l and there are three members of n that meet if not k in m: condition. So

Loop 1 prints: 2,3 Loop 2 prints: 1,3 Loop 3 prints: 1,2 Loop 4 prints: 1,2,3

Answer 4

>>> l 
['1', '2', '3']
>>> n 
['1ab', '2an', '3bca', '4adc']
for el in n: 
 if(el[0] not in l):
 print(el)
4adc

Or if you just want to print 4, based on your list sequence:

for el in n: 
 if(el[0] not in l):
 print(el[0])

Now you just added to your question, "but contains "a"", add second iff.

 for el in n: 
 if(el[0] not in l):
 if('a' in el):
 print(el[0],el)