Increment number in string

Question

I am trying to get the following output until a certain condition is met.

test_1.jpg
test_2.jpg
..
test_50.jpg

The solution (if you could remotely call it that) that I have is


fileCount = 0
while (os.path.exists(dstPath)):
   fileCount += 1
   parts = os.path.splitext(dstPath)
   dstPath = "%s_%d%s" % (parts[0], fileCount, parts[1])

however...this produces the following output.

test_1.jpg
test_1_2.jpg
test_1_2_3.jpg
.....etc

The Question: How do I get change the number in its current place (without appending numbers to the end)?

Ps. I'm using this for a file renaming tool.

UPDATE: Using various ideas below, i've discovered a loop that works


dstPathfmt = "%s_%d%s"
parts = os.path.splitext(dstPath)
fileCount = 0
while (os.path.exists(dstPath)):
   fileCount += 1
   dstPath = parts[0]+"_%d"%fileCount+parts[1]

Michael Mrozek · Accepted Answer · 2010-05-06 05:31:01Z

1

It's probably easiest to keep dstPath something like "test_%d.jpg", and just pass it a varying count:

dstPath = "test_%d.jpg"
i = 1
while os.path.exists(dstPath % i):
    i += 1
dstPath = dstPath % i # Final name

answered May 6, 2010 at 5:31

Michael Mrozek

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1 Comment

iform Over a year ago

a very interesting idea...however i need the flexibility of changing any part of the filename to reflect the original. However, I did change my algorithm above to include a little of both your answer and the answer below. now have dstPathfmt = "%s_%d%s" parts = os.path.splitext(dstPath) fileCounter = 0 #to match the first file "test.jpg" while (os.path.exists(dstPath)): fileCounter += 1 dstPath = parts[0]+"_%d"%fileCount+parts[1]

djna · Accepted Answer · 2010-05-06 05:31:12Z

1

Print out the value of parts[0] each time you go round the loop ... I think you may be surprised,

answered May 6, 2010 at 5:31

djna

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4 Comments

iform Over a year ago

I think i fixed the problem. I moved parts = os.path.splitext(dstPath) outside of the while loop. works perfectly now.

djna Over a year ago

yes, as per Michael's suggestion. You do understand why your version didn't I hope?

iform Over a year ago

i think it is because as each loop executed, dstPath changed (because the previous loop had modified it). Moving it outside used the dstPath before it changed by the loop.

djna Over a year ago

Yes, but how did it change? First time round part[0] is file-1, you then added -2 to that, so next time round part[0] is file-1-2. You thought that fileCount was screwy, but it was fine.

Senseful · Accepted Answer · 2010-05-06 05:32:00Z

0

It seems as though your condition os.path.exists(dstPath) is matching the same renamed file multiple times. So for example, it renames test.jpg to test_1.jpg; then renames test_1.jpg to test_1_2.jpg, etc.

answered May 6, 2010 at 5:32

Senseful

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Comments

Aashish Ranjan · Accepted Answer · 2018-09-04 12:29:25Z

0

for j in range(1,10):
    print("test_{0}.jpg".format(j))

enter image description here

answered Sep 4, 2018 at 12:29

Aashish Ranjan

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1 Comment

Aashish Ranjan Over a year ago

Yes, it renames test.jpg to test_1.jpg; then renames test_1.jpg to test_1_2.jpg, etc. You can see the output of the code by clicking the given link.

wwii · Accepted Answer · 2020-09-26 15:41:04Z

0

Update for Python v3.6+ - using formatted string literals:

for n in range(1,51):
    print(f'test_{n}')

answered Sep 26, 2020 at 15:41

wwii

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