为什么Python要修改循环外的列表？

Question

对于此列表：

current_trace = [[3,5,1,5,7,9,4]]

我运行sliding_tristep()方法，其中包括predict()和window()方法：

def predict(lst):
    print "predicting for", lst
    print "result", max(lst) + 0.0
    return max(lst) + 0.0

def window(lst, n=3):
    for x in range(1, len(lst)+1): # or len(l)+n to continue till the end
        yield(lst[max(0, x-n):x])

def sliding_tristep(full_trace, future_step = 2, window_size = 3):

    for user_trace in full_trace:
        for current_input in window(user_trace):
            counter = 0
            trace = current_input
            accumulator = []
            while counter <= future_step:
                next_prediction = predict(trace)  
                trace.append(next_prediction)
                accumulator.append(next_prediction)
                trace = trace[-window_size:]
                counter += 1

            print current_input, accumulator

当我运行sliding_tristep(current_trace)时，在print current_input, accumulator行的输出中，我注意到current_input已经被修改，尽管它不在while循环中，while循环在sliding_tristep(current_trace)中进行计算。你知道吗

我想知道为什么会这样？python怎么可能修改一个在后续循环中根本没有使用的列表。你知道吗

Answer 1

I run sliding_tristep(current_trace), in the output for the print current_input, accumulator line I notice that the current_trace has been modified

刚刚测试了你的代码：

>>> current_trace = [[3,5,1,5,7,9,4]]
>>> sliding_tristep(current_trace)
...
>>> current_trace
[[3, 5, 1, 5, 7, 9, 4]]

current_trace没有被修改。你知道吗

I wonder why does this happen? How is that possible for python to modify a list which is not used at all in the subsequent loop.

不过，我猜你的意思是current_input，而不是current_trace。你知道吗

current_input被修改，因为trace是对current_input的引用，跟踪被修改。你知道吗

如果要将current_input复制为trace，有一种方法：

>>> foo = [1,2,3]
>>> bar = foo[:]
>>> bar.append(4)
>>> foo
[1, 2, 3]
>>> bar
[1, 2, 3, 4]

应用于您的代码：

def sliding_tristep(full_trace, future_step = 2, window_size = 3):
    for user_trace in full_trace:
        for current_input in window(user_trace):
            counter = 0
            trace = current_input[:] # make a copy of current_input
            accumulator = []
            while counter <= future_step:
                next_prediction = predict(trace)  
                trace.append(next_prediction)
                accumulator.append(next_prediction)
                trace = trace[-window_size:]
                counter += 1                    
            print current_input, accumulator

如果不修改列表中的元素（由于整数是不可变的，所以不能修改），可以按照我在前面的示例中建议的那样进行浅层复制。如果您使用的是可变对象（lists或其他类型的对象），那么您需要使用copy模块进行深度复制。看看这个答案：https://stackoverflow.com/a/184660/1290438关于这个话题。你知道吗

Answer 2

使用trace.extend(current_input)而不是trace = current_input解决了问题。不过，必须事先初始化trace列表。你知道吗

解决方案如下：

def sliding_tristep(full_trace, future_step = 2, window_size = 3):

    for user_trace in full_trace:
        for current_input in window(user_trace): 
            counter = 0
            trace = [] #here is the solution
            trace.extend(current_input) #here is the solution
            accumulator = []
            while counter <= future_step:
                next_prediction = predict(trace)  
                trace.append(next_prediction)
                accumulator.append(next_prediction)
                trace = trace[-window_size:]
                counter += 1