问题:使用Python中的索引向后循环?

我正在尝试从100循环到0。如何在Python中执行此操作?

for i in range (100,0) 不起作用。

I am trying to loop from 100 to 0. How do I do this in Python?

for i in range (100,0) doesn’t work.


回答 0

试试看range(100,-1,-1),第三个参数是要使用的增量(在此处记录)。

此处记录 “范围”选项,开始,停止,步骤)

Try range(100,-1,-1), the 3rd argument being the increment to use (documented here).

(“range” options, start, stop, step are documented here)


回答 1

我认为这是最易读的:

for i in reversed(xrange(101)):
    print i,

In my opinion, this is the most readable:

for i in reversed(xrange(101)):
    print i,

回答 2

for i in range(100, -1, -1)

和一些稍长(且较慢)的解决方案:

for i in reversed(range(101))

for i in range(101)[::-1]
for i in range(100, -1, -1)

and some slightly longer (and slower) solution:

for i in reversed(range(101))

for i in range(101)[::-1]

回答 3

通常在Python中,您可以使用负索引从背面开始:

numbers = [10, 20, 30, 40, 50]
for i in xrange(len(numbers)):
    print numbers[-i - 1]

结果:

50
40
30
20
10

Generally in Python, you can use negative indices to start from the back:

numbers = [10, 20, 30, 40, 50]
for i in xrange(len(numbers)):
    print numbers[-i - 1]

Result:

50
40
30
20
10

回答 4

为什么您的代码不起作用

您的代码for i in range (100, 0)很好,除了

step默认情况下,第三个参数()是+1。因此,必须向range()指定第三个参数才能-1向后退一步。

for i in range(100, -1, -1):
    print(i)

注意:这在输出中包括100&0。

有多种方法。

更好的方法

对于pythonic方式,请检查PEP 0322

这是Python3 pythonic示例,可从100打印到0(包括100和0)。

for i in reversed(range(101)):
    print(i)

Why your code didn’t work

You code for i in range (100, 0) is fine, except

the third parameter (step) is by default +1. So you have to specify 3rd parameter to range() as -1 to step backwards.

for i in range(100, -1, -1):
    print(i)

NOTE: This includes 100 & 0 in the output.

There are multiple ways.

Better Way

For pythonic way, check PEP 0322.

This is Python3 pythonic example to print from 100 to 0 (including 100 & 0).

for i in reversed(range(101)):
    print(i)

回答 5

另一个解决方案:

z = 10
for x in range (z):
   y = z-x
   print y

结果:

10
9
8
7
6
5
4
3
2
1

提示:如果您使用此方法对列表中的索引进行计数,则您希望从’y’值开始为-1,因为列表索引将从0开始。

Another solution:

z = 10
for x in range (z):
   y = z-x
   print y

Result:

10
9
8
7
6
5
4
3
2
1

Tip: If you are using this method to count back indices in a list, you will want to -1 from the ‘y’ value, as your list indices will begin at 0.


回答 6

解决您的问题的简单答案可能是这样的:

for i in range(100):
    k = 100 - i
    print(k)

The simple answer to solve your problem could be like this:

for i in range(100):
    k = 100 - i
    print(k)

回答 7

for var in range(10,-1,-1) 作品

for var in range(10,-1,-1) works


回答 8

简短而甜美。这是我参加codeAcademy类时的解决方案。以rev顺序打印字符串。

def reverse(text):
    string = ""
    for i in range(len(text)-1,-1,-1):
        string += text[i]
    return string    

Short and sweet. This was my solution when doing codeAcademy course. Prints a string in rev order.

def reverse(text):
    string = ""
    for i in range(len(text)-1,-1,-1):
        string += text[i]
    return string    

回答 9

在您的情况下100 - i,您始终可以增加范围并从变量中减去i in range( 0, 101 )

for i in range( 0, 101 ):
    print 100 - i

You can always do increasing range and subtract from a variable in your case 100 - i where i in range( 0, 101 ).

for i in range( 0, 101 ):
    print 100 - i

回答 10

我在一种代码学院练习中尝试过此操作(在不使用reversed或:: -1的情况下反转字符串中的字符)

def reverse(text):
    chars= []
    l = len(text)
    last = l-1
    for i in range (l):
        chars.append(text[last])
        last-=1

    result= ""   
    for c in chars:
        result += c
    return result
print reverse('hola')

I tried this in one of the codeacademy exercises (reversing chars in a string without using reversed nor :: -1)

def reverse(text):
    chars= []
    l = len(text)
    last = l-1
    for i in range (l):
        chars.append(text[last])
        last-=1

    result= ""   
    for c in chars:
        result += c
    return result
print reverse('hola')

回答 11

我想同时向后遍历两个列表,所以我需要负索引。这是我的解决方案:

a= [1,3,4,5,2]
for i in range(-1, -len(a), -1):
    print(i, a[i])

结果:

-1 2
-2 5
-3 4
-4 3
-5 1

I wanted to loop through a two lists backwards at the same time so I needed the negative index. This is my solution:

a= [1,3,4,5,2]
for i in range(-1, -len(a), -1):
    print(i, a[i])

Result:

-1 2
-2 5
-3 4
-4 3
-5 1

回答 12

哦,好吧,我读错了问题,我想这是关于在数组中向后移动吗?如果是这样,我有这个:

array = ["ty", "rogers", "smith", "davis", "tony", "jack", "john", "jill", "harry", "tom", "jane", "hilary", "jackson", "andrew", "george", "rachel"]


counter = 0   

for loop in range(len(array)):
    if loop <= len(array):
        counter = -1
        reverseEngineering = loop + counter
        print(array[reverseEngineering])

Oh okay read the question wrong, I guess it’s about going backward in an array? if so, I have this:

array = ["ty", "rogers", "smith", "davis", "tony", "jack", "john", "jill", "harry", "tom", "jane", "hilary", "jackson", "andrew", "george", "rachel"]


counter = 0   

for loop in range(len(array)):
    if loop <= len(array):
        counter = -1
        reverseEngineering = loop + counter
        print(array[reverseEngineering])

回答 13

您还可以在python中创建自定义反向机制。可以在任何地方用于循环迭代向后

class Reverse:
    """Iterator for looping over a sequence backwards"""
    def __init__(self, seq):
        self.seq = seq
        self.index = len(seq)

    def __iter__(self):
        return self

    def __next__(self):
        if self.index == 0:
            raise StopIteration
        self.index -= 1
        return self.seq[self.index]


>>> d = [1,2,3,4,5]
>>> for i in Reverse(d):
...   print(i)
... 
5
4
3
2
1

You can also create a custom reverse mechanism in python. Which can be use anywhere for looping an iterable backwards

class Reverse:
    """Iterator for looping over a sequence backwards"""
    def __init__(self, seq):
        self.seq = seq
        self.index = len(seq)

    def __iter__(self):
        return self

    def __next__(self):
        if self.index == 0:
            raise StopIteration
        self.index -= 1
        return self.seq[self.index]


>>> d = [1,2,3,4,5]
>>> for i in Reverse(d):
...   print(i)
... 
5
4
3
2
1

回答 14

a = 10
for i in sorted(range(a), reverse=True):
    print i
a = 10
for i in sorted(range(a), reverse=True):
    print i

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