问题
I have code like this:
loopcount = 3
for i in range(1, loopcount)
somestring = '7'
newcount = int(somestring)
loopcount = newcount
so what I want is to modify the range of the for 'inside' the loop.
I wrote this code expecting the range of the for loop would change to (1,7) during the first loop, but it didn't happen.
Instead, no matter what number I put in, it only runs 2 times. (I want 6 times.. in this case)
I checked the value using print like this:
loopcount = 3
for i in range(1, loopcount)
print loopcount
somestring = '7'
newcount = int(somestring)
loopcount = newcount
print loopcount
#output:
3
7
7
7
What is wrong? the number has been changed.
Where is my thinking wrong?
回答1:
The range is created based on the value of loopcount at the time it is called--anything that happens to loopcount afterwards is irrelevant. What you probably want is a while statement:
loopcount = 3
i = 1
while i < loopcount:
somestring = '7'
loopcount = int(somestring)
i += 1
The while tests that the condition i < loopcount is true, and if true, if runs the statements that it contains. In this case, on each pass through the loop, i is increased by 1. Since loopcount is set to 7 on the first time through, the loop will run six times, for i = 1,2,3,4,5 and 6.
Once the condition is false, when i = 7, the while loop ceases to run.
(I don't know what your actual use case is, but you may not need to assign newcount, so I removed that).
回答2:
From the range() docstring:
range([start,] stop[, step]) -> list of integers
Return a list containing an arithmetic progression of integers. range(i, j) returns [i, i+1, i+2, ..., j-1]; start (!) defaults to 0. When step is given, it specifies the increment (or decrement). For example, range(4) returns [0, 1, 2, 3]. The end point is omitted! These are exactly the valid indices for a list of 4 elements.
So, range(1, 10), for example, returns a list like: [1,2,3,4,5,6,7,8,9], so, your code is basically doing:
loopcount = 3
for i in [1, 2]:
somestring = '7'
newcount = int(somestring)
loopcount = newcount
When your for loop is "initialized", the list is created by range().
回答3:
The while-loop answer given by user802500 is likely to be the best solution to your actual problem; however, I think the question as asked has an interesting and instructive answer.
The result of the range() call is a list of consecutive values. The for-loop iterates over that list until it is exhausted.
Here is the key point: You are allowed to mutate the list during iteration.
>>> loopcount = 3
>>> r = range(1, loopcount)
>>> for i in r:
somestring = '7'
newcount = int(somestring)
del r[newcount:]
A practical use of this feature is iterating over tasks in a todo list and allowing some tasks to generate new todos:
for task in tasklist:
newtask = do(task)
if newtask:
tasklist.append(newtask)
回答4:
When the range() function is evaluated in the for-loop it generates a sequence of values (ie a list) that will be used to iterate over.
range() uses the value of loopcount for this. However, once that sequence has been generated, nothing you do inside the loop will change that list, i.e., even if you change loopcount later, the original list will stay the same => the number of iterations will stay the same.
In your case:
loopcount = 3
for i in range(1, loopcount):
becomes
for i in [1, 2]:
So your loop iterates twice, since you have 2 print statements in the loop your get 4 lines of output. Note that you are printing the value of loopcount which is initially 3, but then gets set (and reset) to 7.
If you want to be able to change the iteration number dynamically consider using a while-loop instead. Of course you can always stop/exit any loop early with the use of the break statement.
Also,
somestring = '7'
newcount = int(somestring)
can be simplified to just
newcount = 7
回答5:
To specifically address the question "How do I change the range bounds", you can take advantage of the send method for a generator:
def adjustable_range(start, stop=None, step=None):
if stop is None:
start, stop = 0, start
if step is None: step = 1
i = start
while i < stop:
change_bound = (yield i)
if change_bound is None:
i += step
else:
stop = change_bound
Usage:
myrange = adjustable_range(10)
for i in myrange:
if some_condition:
myrange.send(20) #generator is now bounded at 20
回答6:
It looks like your premise is that you have a default number of times the loop should execute but an occasional condition where it's different. It might be better to use a while loop instead, but regardless you can just do:
if i == some_calculated_threshold:
break
to drop out of the loop instead.
回答7:
You can't increase the number of iterations once the range has been set, but you can break out early, thereby decreasing the number of iterations:
for i in xrange(1, 7):
if i == 2:
break
回答8:
Here is a more complete implementation of the adjustable_range function provided by Joel Cornett.
def adjustable_range(start, stop=None, step=None):
if not isinstance(start, int):
raise TypeError('start')
if stop is None:
start, stop = 0, start
elif not isinstance(stop, int):
raise TypeError('stop')
direction = stop - start
positive, negative = direction > 0, direction < 0
if step is None:
step = +1 if positive else -1
else:
if not isinstance(step, int):
raise TypeError('step')
if positive and step < 0 or negative and step > 0:
raise ValueError('step')
if direction:
valid = (lambda a, b: a < b) if positive else (lambda a, b: a > b)
while valid(start, stop):
message = yield start
if message is not None:
if not isinstance(message, int):
raise ValueError('message')
stop = message
start += step
来源：https://stackoverflow.com/questions/11905606/changing-the-number-of-iterations-in-a-for-loop