Programmieren lernen

s='1232n4jkrbhj4brj4h3278837rbdhjdb5563nnbmnb3333mnb444m4mm6m7m8m' 
print("".join([i if i.isnumeric() else ' ' for i in s]).split())
                

Lösung von: rob ert (tub)

import re
s = "1232n4jkrbhj4brj4h3278837rbdhjdb5563nnbmnb3333mnb444m4mm6m7m8m"
l = [int(x) for x in re.sub(r"[^\d]+", " ", s).split()]
print(l)
                

Lösung von: Name nicht veröffentlicht

function numFilter(str) {
  return str.replace(/\D+/g, ' ').trim().split(' ').map(Number);

}

let string = '1232n4jkrbhj4brj4h3278837rbdhjdb5563nnbmnb3333mnb444m4mm6m7m8m';
console.log(numFilter(string));
                

Lösung von: Lisa Salander (Heidi-Klum-Gymnasium Bottrop)

// NET 6.x | C# 10.x | VS-2022
using static System.Console;
using System.Text.RegularExpressions;

const string str = "1232n4jkrbhj4brj4h3278837rbdhjdb5563nnbmnb3333mnb444m4mm6m7m8m";

// Variante 1 - wie Aufgabenstellung mit Liste als Rückgabewert
WriteLine(string.Join(", ", FilterNums(str)));
static IEnumerable<int> FilterNums(string s) => new Regex(@"\d+").Matches(s).Cast<Match>().Select(x => int.Parse(x.Value));

// Variante 2 - Direktausgabe
WriteLine(string.Join(", ", new Regex(@"\d+").Matches(str)));

// Variante 3 - Direktausgabe oldschool
var n = "";
for (int i = 0; i < str.Length; i++) {
    if (char.IsDigit(str[i])) n += str[i];
    else { if (n.Length > 0) Write(n + " "); n = ""; }
    if (i == str.Length - 1) Write(n);
}
                

Lösung von: Jens Kelm (@JKooP)

// C++ 17
#include <iostream>
#include <vector>
#include <regex>

// Variante 1 - oldschool
std::vector<std::string> filter_nums(std::string str) {
    std::vector<std::string> v{};
    std::string n {};
    for (int i{ 0 }; i < str.length(); i++) {
        if (isdigit(str[i])) n += str[i];
        else { if (n.length() > 0) v.push_back(n); n = ""; }
        if (i == str.length() - 1) v.push_back(n);
    }
    return v;
}

// Variante 2 - Reguläre Ausdrücke
std::vector<std::string> filter_nums_reg(std::string str) {
    std::vector<std::string> v{};
    std::regex reg{ "\\d+" };
    std::sregex_iterator next{ str.begin(), str.end(), reg }, end;
    while (next != end) {
        std::smatch match = *next;
        v.push_back(match.str());
        ++next;
    }
    return v;
}

int main() {
    const std::string str{ "1232n4jkrbhj4brj4h3278837rbdhjdb5563nnbmnb3333mnb444m4mm6m7m8m" };
    const auto fil_reg{ filter_nums_reg(str) };
    for (const auto& f : fil_reg) std::cout << f << " ";
}
                

Lösung von: Jens Kelm (@JKooP)

// NET 6.x | C# 10.x | VS-2022
const string str = "1232n4jkrbhj4brj4h3278837rbdhjdb5563nnbmnb3333mnb444m4mm6m7m8m";

// Variante 4
static string[] FilterNums(string str) => str.Split("abcdefghijklmnopqrstuvwxyz".ToCharArray(), StringSplitOptions.RemoveEmptyEntries);
Console.WriteLine(string.Join(", ", FilterNums(str)));
                

Lösung von: Jens Kelm (@JKooP)

' VBA-Funktionen

Public Function FilterNums(str As String) As Variant

    Dim arr() As Variant
    Dim n As String, m As String
    Dim i As Integer, k As Integer
    
    For i = 1 To Len(str)
        m = Mid(str, i, 1)
        If IsNumeric(m) Then
            n = n & m
        Else
            If Len(n) > 0 Then
                k = k + 1
                ReDim Preserve arr(k)
                arr(k) = n
                n = ""
            End If
        End If
        If i = Len(str) Then
            k = k + 1
            ReDim Preserve arr(k)
            arr(k) = n
        End If
    Next i
    
    FilterNums = arr
    
End Function

' Verweis "VBScript RegularExpressions" einbinden
Public Function FilterNumsReg(str As String) As MatchCollection

    Dim col As MatchCollection
    With New RegExp
        .Pattern = "\d+"
        .Global = True
        Set col = .Execute(str)
    End With

    Set FilterNumsReg = col
    
End Function
                

Lösung von: Jens Kelm (@JKooP)

#include <stdio.h>

int main() {
    char str[] ="1232n4jkrbhj4brj4h3278837rbdhjdb5563nnbmnb3333mnb444m4mm6m7m8m";
    char cmp[] = "abcdefghijklmnopqrstuvwxyz";
    char *tmp;
    printf ("numbers in string:\n", str);
    tmp = strtok (str, cmp);
    while (tmp != NULL) {
        printf ("%s\n", tmp);
        tmp = strtok (NULL, cmp);
    }
}

                

Lösung von: Jens Kelm (@JKooP)