您可以執行以下操作： const countries = [{name: 'Afghanistan',capital: 'Kabul',languages: ['Pashto', 'Uzbek', 'Turkmen'],population: 27657145,flag: 'https://restcountries.eu/data/afg.svg',currency: 'Afghan afghani',},{name: '?land Islands',capital: 'Mariehamn',languages: ['Swedish'],population: 28875,flag: 'https://restcountries.eu/data/ala.svg',currency: 'Euro',},{name: 'Albania',capital: 'Tirana',languages: ['Albanian'],population: 2886026,flag: 'https://restcountries.eu/data/alb.svg',currency: 'Albanian lek',},{name: 'Algeria',capital: 'Algiers',languages: ['Arabic'],population: 40400000,flag: 'https:

您可以使用aggregate： aggregate(cbind(BIP = is.na(BIP), EXP = is.na(EXP)) ~ Time, data = dat, sum)#Also works#with(dat, aggregate(list(BIP = BIP, EXP = EXP), list(Time = Time), function(x) sum(is.na(x)))) Time BIP EXP1 AF 0 12 BF 0 03 DF 1 1 或在dplyr中： library(dplyr)dat %>% group_by(Time) %>% summarise(across(BIP:EXP, ~ sum(is.na(.x)))) Time BIP EXP1 AF 0 12 BF 0 03 DF 1 1

如果不想使用collections包，請創建一個函數來填充字典 c = {}In [23]: for k,v in dict1.items(): ...: for elem in v: ...: if elem not in c: ...: c[elem]=1 ...: else: ...: c[elem]+=1 ...: In [24]: In [24]: c Out[24]: {'Paul': 2, 'Mark': 4, 'Carol': 2, 'Leanne': 4, 'Will': 2, 'Sarai': 1, 'Philip': 2, 'Zach': 2} 如果多個鍵具有最大值 maxVal =0In [44]: for v in c.values(): ...: if v > maxVal: ...: maxVal = v ...: In [45]: maxVal Out[45]: 4 In [31]: keys= []In [34]: for k,v in c.items(): ...: if v == maxVal: ...: keys.append(k) ...: ...: In [36]: print(*keys,sep=',') Mark,Leanne

你可以用pandas做任何事 import pandas as pdimport numpy as npimport stringnp.random.seed(1)df = pd.DataFrame( { "movie": [i for i in string.ascii_uppercase], "vote": np.random.randint(low=10, high=500, size=len(string.ascii_uppercase)) })# If you want a different number change the n_mostn_most = 5df.nlargest(n_most , ["vote"]).plot(kind="bar", x="movie", y="vote", figsize=(15,6), rot=45)

您需要定義元音與其出現頻率之間的映射。在第一步中，計算每個元音出現的次數。在獲得這些信息后，只需按降序對值進行排序，以首先顯示最頻繁的值。下面的代碼在python中，但是這個想法可以用C翻譯： text = input("Enter a text: ")vowels_number = {"a": 0, "e": 0, "i": 0, "o": 0, "u": 0}# Count vowelsfor char in text: if char.lower() in vowels_number: vowels_number[char.lower()] += 1# Sort vowels_number in descending orderprint(f"Most frequent vowels: {sorted(vowels_number.items(), key=lambda item: item[1], reverse=True)}") 對于輸入aaaaAaaeEiIoUu，上述代碼將輸出： Most frequent vowels: [('a', 7), ('e', 2), ('i', 2), ('u', 2), ('o', 1)]

使用basePython；不使用itertools或collections來實現這一點的一種方法是： d = {}mylist = [1,1,2,2,2,3,3,3,3]for i in set(mylist): d[i] = mylist.count(i) 然后，對dict進行排序以確定最頻繁的，等等。 這個答案是出于教育目的發布的，顯示了一種lower-level的方法來完成這個任務，而不需要使用庫。

有很多方法可以解決這個問題。這里有一個：將每個日期映射到只包含年/月/日/小時（即不包含分鐘或秒）的字符串，然后tally結果。然后，您可以查看哪個結果是最常見的（或者您想知道有關聚合數據的任何其他信息）。 dates .map { |date| DateTime.strptime(date, '%m/%d/%Y %H:%M') } .tally

一個可能的解決方案是使用reduce()和forEach()： const myArray = [ ["a", "b", "c"], ["a", "n", "c"], ["e", "f", "g"], ["i", "a", "l"],];const occurrences = myArray.reduce((count, currentValue) => { currentValue.forEach((value) => { count[value] ? ++count[value] : (count[value] = 1); }); return count;}, {});console.log(JSON.stringify(occurrences)); /* StackOverflow snippet: console should overlap rendered HTML area */.as-console-wrapper { max-height: 100% !important; top: 0; } 嚴格地說，上面計算了字母的出現次數，但如果使用count[value] = 0而不是count[value] = 1，則會計算重復的次數。 另一種方法是先使用flat()展平數組，然后使用reduce()，方法同上。同樣，如果您設置count[value] =

首先將current初始化為數組的第一個元素。然后循環其余元素。 當當前元素與current相同時，遞增計數器。當它不同時，重置count和current。 為了得到最高的計數，你需要另一個變量來保存它。無論何時重置count，請檢查它是否高于最高值并將其替換。 并將return語句移出循環。 var arr = ['h', 'a', 'h', 'p', 'h'];function findDuplicate(arr) { arr.sort(); let count = 1; let current = arr[1]; let maxcount = 1; let maxelement = current; for (var i = 1; i < arr.length; i++) { if (arr[i] === current) { count++; } else { if (count > maxcount) { maxcount = count; maxelement = current; } current = arr[i]; count = 1; } } // in case the last element has the highest count if (count > maxcount) { maxcount = count; maxelement

Vanilla JS const data = [ "6", "30", "30", "21", "3", "7", "14", "17", "32", "25", "6", "3", "9", "27", "14", "33", "2", "29", "29", "30", "14", "6", "36", "5", "7", "6", "19", "3", "25", "26", "22", "27", "25", "2", "23", "4", "17", "32", "14", "30", "34", "26", "27", "33", "4", "5", "17", "4", "5", "33", "19", "12", "27", "26", "0", "17", "0", "6", "6", "35", "24", "21", "9", "0", "6", "0", "25", "33", "0", "0", "24", "11", "4", "7", "30", "16", "13", "36", "14", "32", "25", "23", "0", "1", "14", "1", "3", "11", "16", "10", "28", "33", "8", "8", "25", "24", "4", "32", "2", "14", "20", "23",

const s = "Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua";const m = ["a", "em", "i", "el"], x = 3;// given a string s and substring match m, output the occurrences of m in sconst c = (s, m) => [...s].filter((e,i)=>s.substring(i).startsWith(m)).length;// given an array of substring matches, get the occurrences of each substring// and place them in an array as a list of [substring, occurences]// then sort the array by occurrences descending, take the first x entries,// and finally extract only the substring out o

事實上，為了獲得最頻繁的單詞，現有代碼需要稍微修改，以便為重復次數最多的變量提供一個變量，當檢測到更頻繁的單詞時，該變量必須更新。此特定任務不需要額外的陣列/數據結構。 String arr[] = str.split(" ");int maxFreq = 0;String mostRepeated = null;for (int i = 0; i < arr.length; i++) { String temp = arr[i]; int count = 1; for (int j = i + 1; j < arr.length; j++) { if (temp.equalsIgnoreCase(arr[j])) count++; } if (maxFreq < count) { maxFreq = count; mostRepeated = temp; }}System.out.println(mostRepeated + ": " + maxFreq); 對于輸入： String str = "I am he as you are he as you are me and we are all together"; Output: are: 3 更快的實現可能包括將重復值設置為null，以便稍后跳過它們： for (int i = 0; i < arr

以下建議規范： cleanly compiles 執行所需的功能 只包含實際使用的頭文件 現在提議的準則是： #include <stdio.h>void reading_numbers( int Max, int array[ Max ][2]){ for( int i = 0; i < Max; i++ ) { printf("Insert the numbers\n"); scanf("%d", &array[i][0]); array[i][1] = 0; }}void most_present_number( int Max, int array[ Max ][2] ){ for( int i=0; i < Max; i++ ) { for( int j=i; j<Max; j++ ) { if ( array[i][0] == array[j][0] ) { array[i][1]++; } } } int most_present_one = array[0][0]; int most_present_one_counter = array[0][1]; for( int i=1; i<Max; i++ ) { if( most_present_one_counter < array[i][1] ) { most_pr

我覺得你們很接近。像這樣的東西可能有用。切換了map和countBy。首先進行計數，然后映射到結果對象上 const blogs = [{_id: "5a422a851b54a676234d17f7",title: "React patterns",author: "Michael Chan",url: "https://reactpatterns.com/",likes: 7,__v: 0}, { _id: "5a422aa71b54a676234d17f8", title: "Go To Statement Considered Harmful", author: "Edsger W. Dijkstra",url: "http://www.u.arizona.edu/~rubinson/copyright_violations/Go_To_Considered_Harmful.html",likes: 5,__v: 0}, {_id: "5a422b3a1b54a676234d17f9", title: "Canonical string reduction", author: "Edsger W. Dijkstra", url: "http://www.cs.utexas.edu/~EWD/transcriptions/EWD08xx/EWD808.html", likes:

首先，您需要計算一個字符出現的次數。您可以使用reduce方法循環每個字符，并將該字符作為鍵存儲在對象中。鍵的值應該表示字符出現的次數。 它看起來像下面的物體。 { "t": 4, "e": 5, "s": 2} 從這里開始，您需要查找最高的數字并返回與之對應的鍵。我借用了這個SOthread的解決方案，使用另一個reduce循環獲得對象的最高計數。 const string = "Teeeeessttt";const mostOccuringCharacter = string => { // Count the occurence of each character. const count = [...string.toLowerCase()].reduce((counter, char) => { counter[char] = (counter[char] || 0) + 1; return counter; }, {}); // Compare the values with each other and return the // character with the highest count. const highestChar = Object.keys(count).reduce((a, b) => count[a] > count[b