循环里面循环？

Question

I have this code as an example:

for i in func(x):
    count = 1
    while i > 0:
        if count % 2 == 0:
            print("(:")
            count += 2
        else:
            print("(:")
            count += 2

Or this one:

for i in func(x):
    count = 0
    if count % 2 == 0:
        print("(:")
        count += 1
    else:
        print(":(")
        count += 1

They don't work as I want them to work.

The first one just enters in an infinite loop, the second one only prints the happy smiley I want it to print the happy smiley then the sad smiley an so on

This is the original code

Answer 1

It looks like you're trying to use i and count as two different values to use while iterating, and I don't think you need to do this to get what you want.

First of all - a while statement is basically a loop that runs until it is False. So if you have

green = True
while green:
    print("Hello")

...this will never end. Placing the while in a for loop doesn't change that; it simply waits for the while my_condition to turn False.

Additionally, in your first loop you are incrementing count -- but the while loop is looking at the value of i , which never gets a chance to change because the while never becomes untrue!

Finally - in your second loop, you're setting count at the beginning of each iteration! This means that count will never actually progress in a meaningful way; you set it to 0, print an emoticon, increment it by 2, then continue in the for loop - resetting count to 0!

Basically, just realize that you can use i directly, there's no need for a count variable as well. for i in something will iterate over whatever is in something if it is an iterable thing. So for example:

for i in range(0, 10):
    if i % 2 == 0:
        print(":)")
    else:
        print(":(")

EDIT: After seeing the original code, it looks like you are trying to produce a generator object.

Using yield in conjunction with while or for will net you a generator - an object you can iterate over. Here's a baseline example of that.

>>> def baz(x):
...     i = 0
...     while i > x:
...             yield i
...             i += 1
>>> baz(4)
<generator object baz at 0x1004935a0>
>>> me = baz(4)
>>> me.next()
0
>>> me.next()
1
>>> me.next()
2
>>> me.next()
3
>>> me.next()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
StopIteration
>>> 

In a situation where you just want to generate a set number of objects, you can skip the while and just write a for , which will save you a couple of lines. For example, the following code does the same as the above:

def foo(x):
    for i in range(0, x):
        yield i

However, in the example you've shown me there seems to be some more involved math during the while loop, so it does make sense to use a while sometimes. Imagine a situation where you wanted to exit the loop when i became prime, for example…

def until_prime(x):
    while is_not_prime(x): ## or, 'while not is_prime(x)', whatever our imaginary function does
        x = x * (x + 1)
        yield x

I have no idea if the code above will ever generate a prime number, so please don't judge it too harshly. :D Nevertheless, it is an example of a situation where you couldn't necessarily write a for loop, so you'd need to wait until a condition arose.

Answer 2

You probably want the enumerate() function. It appears that i is never used, but that is actually irrelevant.

The enumerate function takes an iterable (so anything you can loop over) and returns a tuple with the count and each item in order.

for index,item in func(x):
    if index % 2 == 0:
        print("(:")
    else:
        print(":(")

This way, even if you don't use i (or in the above case item ) yet , you have the option of doing so in the future.