How to determine whether a string is an integer in Python

Python's method of determining whether a string is an integer: First, you can determine the sign of the integer based on the first character of the string; then traverse the string from right to left, for example, 111, which can be seen as [11 *10 1], and 11 can be seen as [1*10 1].

The operating environment of this tutorial: Windows 7 system, python version 3.9, DELL G3 computer.

Python method to determine whether a string is an integer:

Method 1: Recursive method

For integers, For example, 111 can be seen as 11*10 1, and 11 can be seen as 1*10 1. And -111 can be regarded as (-11)*10-1, and -11 can be regarded as (-1)*10-1. According to this feature, a recursive method can be used to solve the problem, which can be determined based on the first character of the string. The positive and negative of the integer, and then traverse the string from right to left.

class Test:
    def __init__(self):
        self.flag = None
 
    def getFlag(self):
        return self.flag
 
    # 判断c是否是数字，如果是返回数字，且设置flag=True,否则返回False
    def isNumber(self, c):
        return c >= &#39;0&#39; and c <= &#39;9&#39;
 
    # 判断str是否是数字
    def strtoint(self, strs, length):
        if length > 1:
            if not self.isNumber(list(strs)[length - 1]):
                # 不是数字
                print("不是数字")
                self.flag = False
                return -1
 
            if list(strs)[0] == &#39;-&#39;:
                return self.strtoint(strs, length - 1) * 10 - (ord(list[length - 1]) - ord(&#39;0&#39;))
            else:
                return self.strtoint(strs, length - 1) * 10 + ord(list(strs)[length - 1]) - ord(&#39;0&#39;)
 
        else:
            if list(strs)[0] == &#39;-&#39;:
                return 0
            else:
                if not self.isNumber(list(strs)[0]):
                    print("不是数字")
                    self.flag = False
                    return -1
                return ord(list(strs)[0]) - ord(&#39;0&#39;)
 
    def strToint(self, s):
        self.flag = True
        if s == None or len(s) <= 0 or (list(s)[0] == &#39;-&#39; and len(s) == 1):
            print("不是数字")
            self.flag = False
            return -1
 
        if list(s)[0] == &#39;+&#39;:
            return self.strToint(s[1:len(s)], len(s) - 1)
        else:
            return self.strToint(s, len(s))

Method 2: Non-recursive method

First determine the positive or negative of the integer through the value of a character Negativity, then remove the sign bit, treat the following string as a positive number, and return the correct result based on the positivity after the processing is completed. The implementation method is to traverse the string from left to right to calculate the value of the integer. Taking "123" as an example, when traversing to '2', the result is 1*10 2=12, and when traversing to '3', the result is 12*10 3=123.

class Test:
    def __init__(self):
        self.flag = None
 
    def getFlag(self):
        return self.flag
 
    # 判断c是否是数字，如果是返回True,否则返回False
    def isNumber(self, c):
        return c >= "0" and c <= "9"
 
    def strToint(self, strs):
        if strs == None:
            self.flag = False
            print("不是数字")
            return -1
        self.flag = True
 
        res = 0
        i = 0
        minus = False  # 是否是负数
 
        if list(strs)[i] == &#39;-&#39;:  # 结果是负数
            minus = True
            i += 1
 
        if list(strs)[i] == &#39;+&#39;:  # 结果是正数
            i += 1
        while i < len(strs):
            if self.isNumber(list(strs)[i]):
                res = res * 10 + ord(list(strs)[i]) - ord(&#39;0&#39;)
            else:
                self.flag = False
                print("不是数字")
                return -1 
            i += 1
        return -res if minus else res
 
 
if __name__ == "__main__":
    t = Test()
    s = "-543"
    print(t.strToint(s))
    s = "543"
    print(t.strToint(s))
    s = "+543"
    print(t.strToint(s))
    s = "++43"
    result = t.strToint(s)
    if t.getFlag():
        print(result)

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