如何在列表中找到最频繁的

Question

我是python的新手，我需要找到列表中出现次数最多的元素pdInput以及列表中有多少个元素相同mostFreqenNum

    mostFreqenNum = []

    contMostnum = [0]

    

    ContTraining = int(input('How many time You like to Train you input: '))

        for i in range(ContTraining):

            pdInput = int(

                input('Please input your number whatever you want: '))

            mostFreqenNum.append(pdInput)

            for x in mostFreqenNum:

                coutFreqenNum = contMostnum.count(x)

Answer 1

为什么你不使用 python 的内置模块，statistics。您可以像这样使用模块：import statistics### your input codemode = statistics.mode(mostFreqenNum)print(mode)mode() 接收参数列表类型。然后你可以使用 count()。另一个例子，可能像这样：>>> import statistics>>> lists = [2,3,2,2,3,4,5]>>> mode = statistics.mode(lists)>>> print(mode)2>>> lists.count(2)3>>>

Answer 2

我不确定你到底想做什么，但也许这可行：mostFreqenNum = {}contMostnum = 0myList = [1, 2, 3, 2, 4, 3, 2, 3, 5, 3]for i in myList:    if i in mostFreqenNum:        mostFreqenNum[i] += 1    else:        mostFreqenNum[i] = 1        for x in mostFreqenNum:    if mostFreqenNum[x] > contMostnum:        contMostnum = mostFreqenNum[x]        mostFreqKey = x    else:        continueprint(f'Most frequent key, {mostFreqKey}, seen {contMostnum} times.')

Answer 3

def Prediction_Model_v3():alnv3 = [[],[]]inpv3 = int(input('How many time You like to Train you input V3: '))for i in range(inpv3):    pdInpv3 = int(        input('V3 input number whatever you want: '))    alnv3[0].append(pdInpv3)    mdv3 = statistics.mode(alnv3[0])    if(pdInpv3 == mdv3):        alnv3[1].append(str(len(alnv3[1])))    print('numberInput V3: ', alnv3[0])print('Most Frequent number V3 is ', str(mdv3), ':', str(len(alnv3[1])))pdtISv3 = (((inpv3-int(len(alnv3[1])))*100)/inpv3)print('Result of prediction V3 is: ', str(    mdv3), '=', str(pdtISv3), '%')alnv3.clear()return str(pdtISv3)

Answer 4

给定一个值列表inp，您可以找到最常见的值，如下所示：使用collections.Counterfrom collections import Countermost_common = Counter(inp).most_common(1)输出是一个tuple里面(value, count)使用sortedsorted(inp, key=lambda x: inp.count(x), reverse=True)[0]输出是列表中最常见的值使用numpy：# note only works with numeric valuesnp.argmax(np.bincount(inp))输出是列表中最常见的值另一个使用builtins：max(set(inp), key=inp.count)输出是列表中最常见的值另一个使用pandas：import pandas as pdpd.value_counts(inp).index[0]输出是列表中最常见的值