Python 向下枚举或使用自定义步骤

Question

How to make Python's enumerate function to enumerate from bigger numbers to lesser (descending order, decrement, count down)? Or in general, how to use different step increment/decrement in enumerate ?

For example, such function, applied to list ['a', 'b', 'c'] , with start value 10 and step -2 , would produce iterator [(10, 'a'), (8, 'b'), (6, 'c')] .

Answer 1

I haven't found more elegant, idiomatic, and concise way, than to write a simple generator:

def enumerate2(xs, start=0, step=1):
    for x in xs:
        yield (start, x)
        start += step

Examples:

>>> list(enumerate2([1,2,3], 5, -1))
[(5, 1), (4, 2), (3, 3)]
>>> list(enumerate2([1,2,3], 5, -2))
[(5, 1), (3, 2), (1, 3)]

If you don't understand the above code, read What does the "yield" keyword do in Python? and Difference between Python's Generators and Iterators .

Answer 2

One option is to zip your iterable to a range :

for index, item in zip(range(10, 0, -2), ['a', 'b', 'c']):
    ...

This does have the limitation that you need to know how far the range should go (the minimum it should cover - as in my example, excess will be truncated by zip ).

If you don't know, you could roll your own "infinite range " and use that:

>>> def inf_range(start, step):
    """Generator function to provide a never-ending range."""
    while True:
        yield start
        start += step


>>> list(zip(inf_range(10, -2), ['a', 'b', 'c']))
[(10, 'a'), (8, 'b'), (6, 'c')]

Answer 3

Another option is to use itertools.count , which is helpful for "enumerating" by a step , in reverse .

import itertools

counter = itertools.count(10, -2)
[(next(counter), letter) for letter in ["a", "b", "c"]]
# [(10, 'a'), (8, 'b'), (6, 'c')]

Characteristics

• concise
• the step and direction logic is compactly stored in count()
• enumerated indices are iterated with next()
• count() is inherently infinite; useful if the terminal boundary is unknown (see @jonrsharpe)
• the sequence length intrinsically terminates the infinite iterator

Answer 4

Here is an idiomatic way to do that:

list(zip(itertools.count(10,-2), 'abc'))

returns: [(10, 'a'), (8, 'b'), (6, 'c')]

Answer 5

May be not very elegant, using f'strings the following quick solution can be handy

my_list = ['apple', 'banana', 'grapes', 'pear']
p=10
for counter, value in enumerate(my_list):
    print(f" {counter+p}, {value}")
    p+=9

> 10, apple
> 20, banana
> 30, grapes
> 40, pear

Answer 6

If you don't need iterator keeped in variable and just iterate through some container, multiply your index by step.

container = ['a', 'b', 'c']
step = -2

for index, value in enumerate(container):
    print(f'{index * step}, {value}')


>>> 0, a
-2, b
-4, c