Python定义多模态函数

Question

我不需要创建一个可以判断以下情况的函数：

• 没有模式
• 有1模式
• 如果是多模列表

我得到了前2分的掩护：

lista = [1,2,2,3,3,4]
contador = {}

for i in lista:
    cuenta = lista.count(i)
    contador[i] = cuenta

maximo =(0)
moda = [0]

for i in contador:
    if(contador[i]>maximo):
        maximo = contador[i]
        moda = i
        freq = contador[i]


if maximo == 1:
    print("There is no mode")
else:
    print ("Mode is: %d, with a frequency of: %d" % (moda, freq))

但我很难找到一种方法来定义一个列表是否是多式联运的。我想先定义哪个频率是最高的，然后检查contador，取出下面的所有频率，但它似乎行不通：

for i in contador:
    if contador[i] < max:
        delete = [i]
        del contador[delete]

对怎么做有什么想法吗？

谢谢

Answer 1

def moda(x):
    input_string = input("Lista de numeros separados por un espacio: ")
    lista = input_string.split()
    contador = {}
    for i in lista:
        cuenta = lista.count(i)
        contador[i] = cuenta
    maximo =(0)
    for i in contador:
        if(contador[i]>maximo):
            maximo = contador[i]
        modas = {}
        for i in contador:
            if contador[i] == maximo:
                modas[i] = contador[i]
    if maximo == 1:
        return("No hay modas")
    elif len(modas)>1:
        return ("Las modas y sus frecuencias son:")
        return (modas)
    else:
        return ("la moda y su frecuencia es:")
        return (modas)

Answer 2

对于多修饰符，您需要检查最大频率和计数频率计数。如果它是1，那么就意味着没有模式，超过1，也就是1，那么就有1模式，i频率计数大于1，并且有多个值，有相同的计数，那么它是多个数列。

lista = [1,2,3,3,4]

res  = {}
for i in lista:
    if i not in res.keys():
        res[i]=1
    else:
        res[i]+=1
freq = res.values()
max_freq = max(freq)
if max_freq==1:
    print('no mode')
else:
    key = ''
    if list(freq).count(max_freq)==1:
        for k, v in res.items():
            if v==max_freq:
                print('mode is {}, frequency is {}'.format(k, max_freq))
                break
    else:
        print('multimodeal')
        for k, v in res.items():
            if v==max_freq:
                print('mode is {}, frequency is {}'.format(k, max_freq))

Answer 3

对现有代码最简单的修改如下：

maximo = 0
moda = []

for i in contador:
    if(contador[i] > maximo):
        maximo = i
        freq = contador[i]
        moda = []
    if(contador[i]==freq):
        moda.append(i)

并将最后的print更改为：

print ("Mode is: %s, with a frequency of: %d" % (moda, freq))

整个过程通过库函数进行了简化：

from collections import Counter

lista = [1, 2, 2, 3, 3, 4]

contador = Counter(lista)
# get the second part (the count) from the first element of the first most common element
freq = contador.most_common(1)[0][1]
# get all the x's for which the count is freq
moda = [x for x, c in contador.items() if c == freq]

if freq == 1:
    print("There is no mode")
else:
    print("Mode is: %s, with a frequency of: %d" % (moda, freq))