这个函数做什么？

Question

def fun1(a,x):
    z = 0
    for i in range(len(a)):
        if a[i] == x:
            z = z + 1
    return z

Answer 1

计算数组a中x重复元素的数目。在

Answer 2

它计算并返回x在数组a中出现的次数。更广泛地说，a可以是任何可索引的对象。请参见Python Language Reference v2.6.3的5.3.2 Subscriptions：

5.3.2. Subscriptions

A subscription selects an item of a sequence (string, tuple or list) or mapping (dictionary) object:

 subscription ::=  primary "[" expression_list "]"

The primary must evaluate to an object of a sequence or mapping type.

If the primary is a mapping, the expression list must evaluate to an object whose value is one of the keys of the mapping, and the subscription selects the value in the mapping that corresponds to that key. (The expression list is a tuple except if it has exactly one item.)

If the primary is a sequence, the expression (list) must evaluate to a plain integer. If this value is negative, the length of the sequence is added to it (so that, e.g., x[-1] selects the last item of x.) The resulting value must be a nonnegative integer less than the number of items in the sequence, and the subscription selects the item whose index is that value (counting from zero).

A string’s items are characters. A character is not a separate data type but a string of exactly one character.

Answer 3

它计算a中等于x的元素的数量。它假定a是可索引的（如字符串或列表）

def fun1(a,x):              #Defines a function with 2 parameters, a and x
    z = 0                   #Initializes the counter
    for i in range(len(a)): #len(a) returns the length of a, range(len(a)) 
                            #returns an enumerator from 0 to len(a) - 1
        if a[i] == x:       #which is then used here to index a
            z = z + 1       #if the ith element of a is equal to x, increment counter
    return z                #return the counter

给定标题更改后，您可以执行以下函数：

^{pr2}$

或者

> fun1([1,2,3,4,4,3,2,1],4)
2