Python 2.7.2 Win32＃遍历列表，列表的缓冲区部分，多次调用函数，返回相同

Question

I have coded up some small Python script that calls a function multiple times. The function should always return other values, because the buffer is always unique (due to .pop(0) ). But the return is always the same. If I call the function two times like result1 = x(buffer1) and result2 = x(buffer2) outside of the loop, the results are unique. But as soon as I try to call the function like in the Pseudo- Code, the result is always the same. It seems as if Python is executing the function only once, and then printing the result of it. I have implemented this code in PHP and VB.net , and it works as it should. It seems to be a specific Python thing. Can somebody please explain. :/

Pseudo- Code

function x()
    list = []
    buffer = []
    for l in list:
        buffer.append(l)
        if len(buffer) == y:
            return = x(buffer)
        buffer.pop(0)

Python- Code

    ema_length = {66:0,74:0,89:0}
    def ema(length):    
        relevant = len(length)
        multiplier = 2 / (relevant + 1)
        sum_vector = 0
        for vector in length:
            sum_vector += vector    
            current_vector = vector
        sma = sum_vector / relevant
        if ema_length [relevant] == 0:
            ema_length [relevant] = sma 
        else:
            ema_length [relevant] = (current_vector - ema_length [relevant]) * multiplier + ema_length [relevant]
        return ema_length [relevant]

    buffer = []
    for d in data:
        buffer.append(d)
        if len(buffer) == 89:
            ema66 = ema(buffer [89 - 66:89]) # should be unique
            ema74 = ema(buffer [89 - 74:89]) # should be unique
            ema89 = ema(buffer [89 - 89:89]) # should be unique
            ema_overall = ema66 - ema74 - ema89 
            buffer.pop(0)

Answer 1

You are passing your buffer to the function, but are not using it.. Have it like this. :-

def x(old_buffer):
    list = []
    buffer = old_buffer
    for l in list:
        buffer.append(l)
        if len(buffer) == y:
            return = x(buffer)
        buffer.pop(0)

We define function using def not function ...

BTW, what is your purpose of using this -> return = x(buffer) way of return. ?? What is that = doing in between??

Also, I don't see anything added to the list before using it.. Are you pasing any list into that function ?? Else you are re-creating that too on every call..

Answer 2

I think this is the problem:

def ema(length):    
    # ...
    for vector in length:
        sum_vector += vector    
        current_vector = vector
    sma = sum_vector / relevant
    if ema_length [relevant] == 0:
        ema_length [relevant] = sma 
    else:
        ema_length [relevant] = (current_vector - ema_length [relevant]) * multiplier + ema_length [relevant]
    return ema_length [relevant]

The function ema gets a list of values from your buffer. The last element is always the same (as you always call it with buffer[89 - X:89] ). Now, you loop through the elements, sum them up and set current_vector to the current one. After the loop finished, current_vector is set to the last one, buffer[89] —always. Afterwards you use current_vector to set up the return value. So the problem is that you exit the loop to early.

Answer 3

multiplier = 2 / (relevant + 1) # floats are rounded! => * with 0 => same result
multiplier = float(2) / float(relevant + 1) # unique values! :)