Python定义多模态函数

我不需要创建一个函数来判断：

没有模式 有1种模式 如果是多式联运清单 我得到了前2个积分：

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))

但我正在努力找到一种方法来定义列表是否是多模态的。我想先定义哪个频率是最高的，然后检查contador以删除下面的所有频率，但它似乎不起作用：

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

有什么办法吗

---

谢谢

对于多式联运，您需要检查最大频率并计算频率计数。如果为1，则表示无模式，大于1，正好为1，则有1个模式，且i频率计数大于1，且具有多个值，且计数相同，则为多峰

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))

---

对于多式联运，您需要检查最大频率并计算频率计数。如果为1，则表示无模式，大于1，正好为1，则有1个模式，且i频率计数大于1，且具有多个值，且计数相同，则为多峰

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))

---

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)

并将最终打印更改为：

整个过程通过库函数简化：

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))

---

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)

并将最终打印更改为：

整个过程通过库函数简化：

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))