如何在 Python 的循环中更改 for-loop 迭代器变量？

Question

I want to know if is it possible to change the value of the iterator in its for-loop?

For example I want to write a program to calculate prime factor of a number in the below way :

def primeFactors(number):
    for i in range(2,number+1):
        if (number%i==0)
            print(i,end=',')
            number=number/i 
            i=i-1 #to check that factor again!

My question : Is it possible to change the last two line in a way that when I change i and number in the if block, their value change in the for loop!

Update: Defining the iterator as a global variable, could help me? Why?

Answer 1

No.

Python's for loop is like other languages' foreach loops. Your i variable is not a counter, it is the value of each element in a list, in this case the list of numbers between 2 and number+1. Even if you changed the value, that would not change what was the next element in that list.

Answer 2

Short answer (like Daniel Roseman's): No

Long answer: No, but this does what you want:

def redo_range(start, end):
    while start < end:
        start += 1
        redo = (yield start)
        if redo:
            start -= 2

redone_5 = False
r = redo_range(2, 10)
for i in r:
    print(i)
    if i == 5 and not redone_5:
        r.send(True)
        redone_5 = True

Output:

3
4
5
5
6
7
8
9
10

As you can see, 5 gets repeated. It used a generator function which allows the last value of the index variable to be repeated. There are simpler methods ( while loops, list of values to check, etc.) but this one matches you code the closest.

Answer 3

The standard way of dealing with this is to completely exhaust the divisions by i in the body of the for loop itself:

def primeFactors(number):
    for i in range(2,number+1):
        while number % i == 0:
            print(i, end=',')
            number /= i

It's slightly more efficient to do the division and remainder in one step:

def primeFactors(number):
    for i in range(2, number+1):
        while True:
            q, r = divmod(number, i)
            if r != 0:
                break
            print(i, end=',')
            number = q

Answer 4

The only way to change the next value yielded is to somehow tell the iterable what the next value to yield should be . With a lot of standard iterables, this isn't possible . however, you can do it with a specially coded generator:

def crazy_iter(iterable):
  iterable = iter(iterable)
  for item in iterable:
    sent = yield item
    if sent is not None:
      yield None  # Return value of `iterable.send(...)`
      yield sent

num = 10
iterable = crazy_iter(range(2, 11))
for i in iterable:
  if not num%i:
    print i
    num /= i
    if i > 2:
      iterable.send(i-1)

I would definitely not argue that this is easier to read than the equivalent while loop, but it does demonstrate sending stuff to a generator which may gain your team points at your next local programming trivia night.

Answer 5

It is not possible the way you are doing it. The for loop variable can be changed inside each loop iteration, like this:

for a in range (1, 6):
    print a
    a = a + 1
    print a
    print

The resulting output is:

1
2

2
3

3
4

4
5

5
6

It does get modified inside each for loop iteration.

The reason for the behavior displayed by Python's for loop is that, at the beginning of each iteration, the for loop variable is assinged the next unused value from the specified iterator . Therefore, whatever changes you make to the for loop variable get effectively destroyed at the beginning of each iteration.

To achieve what I think you may be needing, you should probably use a while loop , providing your own counter variable, your own increment code and any special case modifications for it you may need inside your loop. Example:

a = 1
while a <= 5:
    print a
    if a == 3:
        a = a + 1
    a = a + 1
    print a
    print

The resulting output is:

1
2

2
3

3
5

5
6

Answer 6

Yes, we can only if we dont change the reference of the object that we are using. If we can edit the number by accessing the reference of number variable, then what you asked is possible.

A simple example:

a=[1,2,3]
a=a+[4]==>here, a new object is created which plots to different address.
a+=[4]==>here , the same object is getting updated which give us the desired result.



number=10
list1=list(range(2,number+1))
# list1   
for i in list1:
    print(list1,i)
    if (number%i==0):
        print(i,end=',')
        number=number//i #we can simply replace it with number//=i to edit the number without changing the reference or without creating a new object.
        try:
            [list1.pop() for i in range(10,0,-1) if(i>number)]
             #here pop() method is working on the same object which list created by number refers. so, we can able to change the iterable in the forloop. 

        except:
            continue
        i=i-1 #to check that factor again!