Question

我试着做这两个功能：

# majority(x)
# pre-condition: x is a list of booleans (True or False values)
# post-condition: returns True if there are strictly more Trues than False in x; otherwise returns False.

def majority(x):
    numTrue = 0
    numFalse = 0
    for i in x:
        if (i==True):#check to see how many Trues there are 
            (numTrue) += 1
        else:# check to see how many Falses there are
            (numFalse) += 1
        if ((numTrue) > (numFlase)):# check to see what is the majority of the list(trues or falses)
            return True
        else:
            return False

在这一部分中，它向我显示numFalse未定义，我不知道原因：

# expletive_deleted(x)
# pre-condition: x is a list of strings
# post-condition: replace each string of length 4 in x with the string ****.
#return nothing.

def expletive_deleted(x):
    for n,i in enumerate(x):#enumerate works like 2 for loops together 
        if (len(x[n])==4):
            x[n] = '****'

并且每当我尝试测试该功能时，它都没有显示任何内容。

Answer 1

您在第一个numFalse区块numFlase中拼错了if。

至于第二个，你实际上并没有打印/做任何事情。假设您从其他地方调用该函数，则需要根据您的要求打印或返回该值。另外，在您的情况下，您可以说if len(i) == 4，因为x[n]等同于i（n是索引，i是值）。

Answer 2

def majority(lst):
    return sum(1 for b in lst if b) > (len(lst) // 2)

def expletive_deleted(lst): #note: not inplace as required
    return [s if len(s) != 4 else "****" for s in lst]

就地：

def expletive_deleted(lst):
    for i, s in enumerate(lst):
        if len(s) == 4:
           lst[i] = "****"

Answer 3

试试这个：

def majority(x):
    return len([y for y in x if y]) > len(x) / 2

和

def expletive_deleted(x):                                             
    replacer =  lambda s: len(s) == 4 and "****" or s                 
    return map(replacer, x)

most和expletive_deleted python函数

3 个答案: