Question

这是我的代码：

def L_value_Change(k):
    global L
    L = k
    return L

def  applyF_filterG(L, f, g):
    L
    k = []
    for i in L:
       if g(f(i)):
          k.append(i)
    L = k
    L_value_Change(k)
    if L == []:
        return -1
    else :
        return max(L)

当我输入此代码时，评分者告诉我这是不正确的！所以我读了测验的介绍，导师写道“全局变量不起作用”。如何在不使用单词L的情况下使用函数更改global变量？如果你尝试我的代码并给它所需的输入，它会给你一个正确的答案，但是评分者告诉我这是错误的。

Answer 1

如果要将全局变量重新绑定到其他对象，则需要{ "from": 0, "size": 20, "query": { "function_score": { "query": { "bool": { "filter": [ { "bool": { "should": { "terms": { "categories.category1Id": [ 63 ] } } } } ] } }, "functions": [ { "gauss": { "updatedDate": { "origin": "2016-10-03 05:10:18", "scale": "0.5h", "decay": 0.1, "offset": "1h" } } }, { "filter": { "term": { "productQuality": "EXCELLENT" } }, "weight": 7 }, { "filter": { "term": { "productQuality": "HIGH" } }, "weight": 5 }, { "filter": { "term": { "productQuality": "MEDIUM" } }, "weight": 3 }, { "filter": { "term": { "productQuality": "LOW" } }, "weight": 1 } ], "score_mode": "sum" } } }关键字。但是如果您想要做的就是更改可变对象，则不需要它。在您的情况下，global是一个列表，可以使用切片操作L进行调整。为了证明：

L[:] = k

Answer 2

列表是可变对象，因此，要更改它们，您只需将它们作为参数传递给函数即可。

def f(i):
    return i + 2

def g(i):
    return i > 5

l = [0, -10, 5, 6, -4]
def applyF_filterG(L, f, g):
    for val in L[:]:
        if not g(f(val)):
            L.remove(val)
    return -1 if not L else max(L)      

print(l) # [0, -10, 5, 6, -4]
applyF_filterG(l, f, g) # Return 6
print(l) # [5, 6]

Answer 3

这是我的代码，它避免了你试图迭代的列表上的变异。

def applyF_filterG(L,f,g):
        """
        Assumes L is a list of integers
        Assume functions f and g are defined for you. 
        f takes in an integer, applies a function, returns another integer 
        g takes in an integer, applies a Boolean function, 
            returns either True or False
        Mutates L such that, for each element i originally in L, L contains  
            i if g(f(i)) returns True, and no other elements
        Returns the largest element in the mutated L or -1 if the list is empty
        """
        # Applying the functions & Mutating L
        i = len(L)-1
        largestnum=0  
        if (len(L)==0):
            return -1
        else:
            while i>=0:
                if not g(f(L[i])):
                   del L[i]
                i-=1    

        #Finding the largest number
        if not L:
            return -1
        if (len(L)==1):
            return L[0]
        else:
            for num in range(len(L)-1):
                if (L[num]>L[num+1]):
                    largestnum=L[num]
                else:
                    largestnum=L[num+1]
            return largestnum

在不使用global关键字的情况下更改函数中的全局变量

3 个答案: