Stack Overflow Asked by Sivaram on December 25, 2021
a = 6
item_list = [1,2,3,4,5]
for items in item_list:
if items == a:
print("match found")
else:
print ("no match")
In this code I want "no match" to be printed only once after completing all the iterations; not on every iteration. How can I modify the code?
You could use a while
loop:
a = 6
item_list = [1, 2, 3, 4, 5]
i = 0
found = False
while True:
i += 1
if item_list[i] == 6:
print("Found")
found = True
if i > len(item_list):
break
if not found:
print("Not found")
Note that 6 is not in the list, so it will print
"Not found"
Answered by Psychzander on December 25, 2021
As others have pointed out, the most efficient solution in this situation is to use an in
test, and no explicit loop is needed.
However, in a more general situation where one is looping until a match is found (if more complex logic means that there is no simple equivalent of in
available), it is worth remembering that a for
loop can have an else
block which is run after the loop completes if it is has not been exited using break
. Taking this code as an example (even though not necessary in this situation), you could do:
a = 6
item_list = [1,2,3,4,5]
for item in item_list:
if item == a:
print("match found")
break
else:
print ("no match")
Note that this is not exactly equivalent to the code in the question if more than one match is found, as here "match found" is only printed once and then the loop is exited via break
. If you were looking for potentially more than one match, for example saving these to a list, then you would not use break
and else
, but would instead test afterwards whether your output list was empty before printing "no match".
Answered by alani on December 25, 2021
a = 6
item_list = [1,2,3,4,5]
found_in_list_boolean = False
for items in item_list:
if items == a:
print("match found")
'''Set the boolean to True as you reached the match found statement '''
found_in_list_boolean = True
'''2 ways from here'''
'''First Way : If found_in_list_boolean is true print something and for false print not found'''
if (found_in_list_boolean):
print("Found")
else:
print("not found")
'''Second Way :If found_in_list_boolean is false and print'''
if (found_in_list_boolean==False):
print("not Found")
Answered by Rahul Shyokand on December 25, 2021
a in b
is generally a better solution, but if you want to loop through the entire list you could create a variable to store
matches in and check if the variable is populated when the for-loop expires.
a = 6
memory = []
item_list = [1,2,3,4,5]
for items in item_list:
if items == a:
memory.append(a)
print("match found")
if len(memory) == 0:
print("No matches")
a
is never gonna show up in the loop since you're only looping over elements in item_list, fyi
Answered by NotActuallyErik on December 25, 2021
Use instead:
a = 6
item_list = [1,2,3,4,5]
if a in item_list:
print("match found")
else:
print ("no match")
No need to iterate list by yourself, use built-in method.
Answered by ipj on December 25, 2021
Instead of looping through, you can use index
on list
or using in
operator
In [39]: a = 6
...: item_list = [1,2,3,4,5]
In [40]: try:
...: index = item_list.index(a)
...: print("match found")
...: except:
...: print("no match")
...:
no match
In [41]: if a in item_list:
...: print("match found")
...: else:
...: print("no match")
...:
no match
Answered by bigbounty on December 25, 2021
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