Using for loop iteration with If statement

Question

I have a list of multiple items. Need to create a loop that finds an item from the list and prints it out. If item not found prints out only once that it hasn't been found.

 for x in range(len(regs)):
    rex = regs[x]
    if re.match(rex,hostname):
        print (dev[x],'Regex matched')
        break
    else:
        print('wrong format')

Currently it prints out multiple times that it hasn't been found. When it matches the regex it prints out at the end. I only want the If Else to prints "wrong format" once.

currently prints out wrong format wrong format wrong format wrong format wrong format wrong format wrong format "DeviceX Regex matched"

a simple solution will be to use a flag

Andy_101
– Andy_101

2019-09-13 10:32:47 +00:00
Commented Sep 13, 2019 at 10:32 — Andy_101
– Andy_101, Commented Sep 13, 2019 at 10:32

Dipen Dadhaniya · Accepted Answer · 2019-09-13 10:39:54Z

1

Use else with for.

As you are breaking out of the for loop as soon as you find one item, the below code will work fine:

for x in range(len(regs)):
    rex = regs[x]
    if re.match(rex,hostname):
        print (dev[x],'Regex matched')
        break
else:
    # Does not execute only when a break statement
    # is executed inside the for loop.
    print('wrong format')

edited Sep 13, 2019 at 10:39

answered Sep 13, 2019 at 10:35

Dipen Dadhaniya

4,6602 gold badges19 silver badges26 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Andy_101 Over a year ago

just a gentle reminder.. it's not a good practice to use for-else construct but in this case its fine I guess

Dipen Dadhaniya Over a year ago

Agreed! But in this short and simple code, I think it is fine.

Nico Griffioen · Accepted Answer · 2019-09-13 10:34:41Z

0

It is printed multiple times, because there are multiple times where the current value does not match. A simple way to fix this is to use a flag, like so:

value_found = False 

for x in range(len(regs)):
    rex = regs[x]
    if re.match(rex,hostname):
        print (dev[x],'Regex matched')
        value_found = True
        break

if not value_found:
    print('wrong_format')

answered Sep 13, 2019 at 10:34

Nico Griffioen

5,4362 gold badges29 silver badges36 bronze badges

Comments

Chinni · Accepted Answer · 2019-09-13 10:35:23Z

0

You can maintain a flag variable as follows.

matched = False
for x in range(len(regs)):
    rex = regs[x]
    if re.match(rex, hostname):
        print(dev[x], "Regex matched!")
        matched = True
        break
if not matched:
    print("Wrong format")

answered Sep 13, 2019 at 10:35

Chinni

1,2901 gold badge16 silver badges28 bronze badges

Comments

Yannick Guéhenneux · Accepted Answer · 2019-09-13 10:44:09Z

0

I think you can make it shorter using comprehensive list such as:

if len([r for r in regs if re.match(r, hostname)]) == 1:
  print(dev[regs.index([r for r in regs if re.match(r, hostname)][0])], 'Regex matched')
else:
  print('not found')

answered Sep 13, 2019 at 10:44

Yannick Guéhenneux

4834 silver badges13 bronze badges

1 Comment

Kins Over a year ago

This code does not work if multiple elements match. The first condition should be len(...) >= 1. And you would probably also need to change the statement in the print.

Chinni · Accepted Answer · 2019-09-13 12:47:41Z

0

This should solve your problem.

flag = 0     
for x in range(len(regs)):
    rex = regs[x]
    if re.match(rex, hostname):
        print(dev[x], 'Regex matched')
        flag = 1
        break
if flag == 0:
    print('wrong format')

edited Sep 13, 2019 at 12:47

Chinni

1,2901 gold badge16 silver badges28 bronze badges

answered Sep 13, 2019 at 10:35

Andy_101

1,30810 silver badges21 bronze badges

Collectives™ on Stack Overflow

Using for loop iteration with If statement

5 Answers 5

2 Comments

Comments

Comments

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

2 Comments

Comments

Comments

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related