从另一个函数内的python函数访问变量值

Question

I have one function that calls for the function rlEncode which is supposed to take the data list and compress it so it counts how many values in a row there are and would output for example [1, 5, 3, 2, 5, 6] and so on. But when I run it just out puts the [1,5] again and again and not moving the n value over however so many spaces in the list. How would I get the n value from the function rlEncode to be used in the other function?

 def rlEncode(n, z, data_list): while data_list[n] == data_list[n+1]: z = z + 1 n = n + 1 while data_list[n] != data_list[n+1]: return n return z def unitTest( ): c = 0 n = 0 z = 1 data_list = [1,1,1,1,1,3,3,5,5,5,5,5,5,6,8,8,1,1,1,5,5,5,5,13,14, 14] compress_list = [ ] while c < (len(data_list)): n = rlEncode(n, 1, data_list) z = rlEncode(0, z, data_list) rlEncode(0, 1, data_list) compress = [data_list[n], z] c = c + 1 compress_list = compress_list + compress print(compress_list) n = n+1

Answer 1

Python passes immutable objects by value. See this previous answer: How do I pass a variable by reference?

The simplest solution in your case is to have the inner function return the value of n to the outer function, which assigns it to its local n .

compress is a list, which is mutable, so you can use += to mutate the list in place rather than creating a new local variable.

I also added a check against the list length, otherwise the references to n+1 will cause IndexError.

I also don't think you need the second while loop in rlEncode but I'll leave that up to you to sort out... :)

def rlEncode(n, z, data_list, compress):
    while (n < len(data_list)-1) and (data_list[n] == data_list[n+1]):
        z = z + 1
        n = n + 1
    while (n < len(data_list)-1) and (data_list[n] != data_list[n+1]):
        compress += [data_list[n], z]
        n = n + 1
    return n

def unitTest(data_list):
    c = 0
    n = 0
    compress = []
    while c < (len(data_list)):
        n = rlEncode(n, 1, data_list, compress)
        c = c + 1
    return ('list: ', data_list, "compressed list: ", compress)

sample_data = [1,1,1,1,1,3,3,5,5,5,5,5,5,6,8,8,1,1,1,5,5,5,5,13, 14, 14]
unitTest(sample_data)

Answer 2

I like doing these things recursively. Python isn't great with recursion but I wanted to see how it's done so I guess I'll share:

def compress(lst):
    if not lst:
        return []
    current = lst[0]
    result = compress(lst[1:]) 

    if not result:
        # item is last
        return [(1, current)]
    nxt = result[0][1]

    if current == nxt:
        # items is same as next
        return [(result[0][0] + 1, current)] + result[1:]

    # different items
    return [(1, current)] + result

To use it:

print [x[0] for x in compress(lst)]

The obvious and efficient way would be a generator:

def group_gen(lst):
    buff = []
    for item in lst:
        if buff and item == buff[0]:
            buff.append(item)
        else:
            if buff:
                yield buff
            buff = [item]
    if buff:
        yield buff

print list(map(len, group_gen(lst)))

Check it using:

print list(map(len, group_gen(lst)))