Python：查找包含通配符的字符串中是否存在子字符串

如何解决Python：查找包含通配符的字符串中是否存在子字符串

我想创建一个接受两个输入 fullstring 和 substring 的函数。如果完整字符串中存在子字符串，则函数应返回 True。否则，它将返回 False。如果子字符串包含通配符 (*)，则通配符可以表示任何单个字符。

例如：

arg1: fullstring = "hitherehello"
arg2: substring = "the*e
output: True

我尝试过的： 下面是一个将子字符串与父字符串匹配的函数，但我似乎无法弄清楚如何集成通配符。

def count_substring(string,sub_string):
    len_ss = len(sub_string)
    for i in range(len(string) - len_ss + 1):
        if string[i:i+len_ss] == sub_string:
            return True

    return False

约束： 我不能使用正则表达式或内置的 Python 函数，例如 find。

解决方法

您可以查找字符串中的每个“通配符”部分，返回位置并随后检查是否有后续：

def count_substring(string,sub_string,lastIndex):
    len_ss = len(sub_string)
    for i in range(len(string) - len_ss + 1):
        if string[i:i + len_ss] == sub_string and lastIndex < i + len_ss:
            return i + len_ss
    return -1


def count_wrapper(string,index):
    positions = []
    for wild_sub_string in sub_string.split('*'):
        index = count_substring(string,wild_sub_string,index)
        positions.append(index)
    # check pairwise
    return all([x + 2 == y for x,y in zip(*[iter(positions)] * 2)])


print(count_wrapper("there","the*e",0))
print(count_wrapper("thera",0))
print(count_wrapper("theresa","the*e*a",0))

出：

True
False
True

,

import java.util.*;
public class ArrayStatistics {

public static void main(String[] args) {
    double total = 0;
    
    Scanner input = new Scanner(System.in);
    
    System.out.print("Enter the size of your array >> ");
    int size = input.nextInt();
    double[] myArray = new double[size];
    System.out.print("Enter the integer values >> ");
    for (int i=0; i<size; i++) {
        myArray[i] = input.nextInt();
        }
    System.out.println("\nIntegers:");
    for (int i=0; i<size; i++) {
        System.out.println(myArray[i]);
        }
    
    double mean = calculateMean(myArray);
    System.out.println("\nMean: " + mean);
    
    double mode = calculateMode(myArray);
    System.out.println("Mode: " + mode);
    
    double median = calculateMedian(myArray);
    System.out.println("Median: " + median);
    
    double SD = calculateSD(myArray);
    System.out.format("Standard Deviation: %.6f",SD);
    }

public static double calculateMean(double myArray[]) {
    int sum = 0;
    
    for(int i = 0; i<myArray.length; i++) {
        sum = (int) (sum + myArray[i]);
    }
    double mean = ((double) sum) / (double)myArray.length;
    
    return mean;
}

public static double calculateMode(double myArray[]) {
    int modeCount = 0;
    int mode = 0;
    int currCount = 0;
    
    for(double candidateMode : myArray) {
        currCount = 0;
        for(double element : myArray) {
            if(candidateMode == element) {
                currCount++;
            }
        }
        if(currCount > modeCount) {
            modeCount = currCount;
            mode = (int) candidateMode;
        }
    }
    return mode;
}

public static double calculateMedian(double myArray[]) {
    Arrays.sort(myArray);
    
    int val = myArray.length/2;
    double median = ((myArray[val]+myArray[val-1])/2.0);
    
    return median;
}

public static double calculateSD(double myArray[]) {
    double sum = 0.0;
    double standardDeviation = 0.0;
    int length = myArray.length;
    for(double num : myArray) {
        sum += num;
        }
    double mean = sum/length;
    for(double num : myArray) {
        standardDeviation += Math.pow(num - mean,2);
        }
    return Math.sqrt(standardDeviation/length);
    }

打印：

def search(fullstring,substring):
    def check(s1,s2):
        for a,b in zip(s1,s2):
            if a != b and b != "*":
                return False
        return True

    for i in range(len(fullstring) - len(substring) + 1):
        if check(fullstring[i : i + len(substring)],substring):
            return True

    return False


print(search("hitherehello","the*e"))

---

更多测试：

True

,

这似乎对我有用：

def check_all_substrings(index,string,substrings):
    for ss in substrings:
        if string[index:index+len(ss)] != ss:
            return False

        # If we matched,move the index along by the length of the substring + 1,so we skip a character
        index = index + len(ss) + 1
    return True

def match(string,substring):
    i = 0
    substrings = substring.split('*')

    while (i < (len(string) - len(substring))):

        if check_all_substrings(i,substrings):
            return True

        i += 1

    return False

# match substring
assert match("hitherehello","there")

#match substring with 1 wild card
assert match("hitherehello","the*e") 

# match 
assert match("hithat","h*t*a")

# wild cards should match exactly 1 character
assert not match("hitherrrrehello","the*e")

# do not match invalid substrings
assert not match("hithat","the*e")
assert not match("hithat","there")