CODEWITHSUNDEEP
python
David Emmerich
David Emmerich

Reputation: 45

How do I create a loop that continues until the original value is repeated?

I'm reading a series of values, like 4, 5, 5, 6, 6, 6, 7, 6, 4, 5, 6, 7, 6, 5

The first value will always be 4 and others will always be >4 until another 4 is reached. I want to continue the loop, but stop before the next 4 is reached.

So my desired output would be: 4, 5, 5, 6, 6, 6, 7, 6

for value in list:
    print(value)

I'm sure it's simple, but I can't think of the logic to stop the loop

for value in list:
    print(value)
    
    if value == 4:
       break

If I do this, it stops at the first 4 of course. How do I get the logic to ignore the first 4 and only apply that if value == 4, after the first loop cycle?

Upvotes: 0

Views: 106

Answers (9)

pintert3
pintert3

Reputation: 117

Use a counter:

num_4s = 0
for value in list:
    if value == 4:
       num_4s += 1
    
    if num_4s > 1:
       break

    print(value)

This would probably even help someone who has to count more than one original value before stopping.

Upvotes: 1

dmlicht
dmlicht

Reputation: 2438

If your stopping condition don't depend on the first element of the list, you can just skip it.

print(list[0])
for value in list[1:]:    
    if value == 4:
       break
    print(value)

If for example, you wanted to design this in a way where your stopping condition depends on the first item, you can do:

stop_value = list[0]
print(stop_value)
for value in list[1:]:    
    if value == stop_value:
       break
    print(value)

Upvotes: 0

Deepak Tripathi
Deepak Tripathi

Reputation: 3233

I think this can be done without for loop like this :

l =  [4, 5, 5, 6, 6, 6, 7, 6, 4, 5, 6, 7, 6, 5]
print(l[:l[1:].index(4) +1])

Upvotes: 0

aasoo
aasoo

Reputation: 114

I suppose there are multiple options, which one is 'correct' would depend on your exact use case. You could skip the first element within your loop:

print(data[0])
for value in data[1:]:    
    if value == 4:
       break

    print(value)

You could also add an additional check for the index:

for i, value in enumerate(data):
    if value == 4 and i > 0:
       break

    print(value)

Upvotes: 1

sandes
sandes

Reputation: 2267

another solution

l = [4, 5, 5, 6, 6, 6, 7, 6, 4, 5, 6, 7, 6, 5]

f = None
idx = 0

while l[idx] not in (f,):
    f = l[0]
    print (l[idx])
    idx += 1

Upvotes: 1

Dejene T.
Dejene T.

Reputation: 989

This is the solution you may help you.

n = [4, 5, 5, 6, 6, 6, 7, 6, 4, 5, 6, 7, 6, 5]
new = [4]
for i in range(1, len(n)):
    
    if n[i] != 4:
        new.append(n[i])
    else:
        break
print(new)

Upvotes: 0

Jean-François Fabre
Jean-François Fabre

Reputation: 140168

simplest logic I could figure out. Use a flag, set it if you meet the value, but first check if already set.

lst = [4, 5, 5, 6, 6, 6, 7, 6, 4, 5, 6, 7, 6, 5]

four_met = False
for e in lst:
    if e == 4:
        if four_met:
            break
        four_met = True
    print(e)

Upvotes: 2

Jortega
Jortega

Reputation: 3790

I would try to use the index of the list.

some_list = [4, 5, 5, 6, 6, 6, 7, 6, 4, 5, 6, 7, 6, 5]
for value in range(len(some_list)):
    print(some_list[value])

    if some_list[value + 1] == 4:
        break

Upvotes: 1

Green Cloak Guy
Green Cloak Guy

Reputation: 24691

Set a flag.

found_4 = False
for value in list:
    print(value)
    
    if value == 4:
       if not found_4:
           found_4 = True
       else:
           break

If you need to stop at, say, the third occurrence of a 4, then you can make the flag an integer instead and count up to it.

Upvotes: 0

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