How to enumerate through list with the first “index” reported as 1? (Python 2.4)
-
06-03-2021 - |
Вопрос
I need my counter to start at 1. Right now I have
for(counter, file) in enumerate(files):
counter += 1
//do stuff with file and counter
But there must be a better way, in Python v2.4
Решение
Instead of counter += 1
, maybe use counter + 1
where you've used counter
.
Alternatively:
for counter, file in ((i + 1, f) for i, f in enumerate(files)):
...
(Python 2.6 and later has some great stuff. Try to upgrade if you can.)
Другие советы
Generators are perfect for this:
def altenumerate( it ):
return ((idx+1, value) for idx, value in enumerate(it))
A simplified for older versions of python:
def altenumerateOld( it ):
idx = 1
for value in it:
yield (idx, value)
idx += 1
You can make your own version of enumerate()
:
def enumerate_1based(iterable):
for index, item in enumerate(iterable):
yield index+1, item
Alternately, add a start
argument, to make it work just like later versions of enumerate()
.
You can use zip()
:
>>> enums = zip(range(1, len(files) + 1), files)
>>> for index, val in enums:
print index, val
I did this like this:
#Emulate enumerate() with start parameter (introduced in Python 2.6)
for i,v in (i+start,v for i,v in enumerate(seq)):
//do stuff
Basically, this is the same, yet a self-contained construct.
for counter, item in enumerate(testlist):
print(counter+1)
print(item)
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