Question

我在这个比较函数中遇到了一些尴尬的代码：

def compare(this, that, encoding="utf-8"):

      if type(this) == type(str()):
           this = this.encode(encoding)
      if type(this) == type(bytes()):
           this = this.decode().encode(encoding)

      if type(that) == type(str()):
           that = that.encode(encoding)
      if type(that) == type(bytes()):
           that = that.decode().encode(encoding)

      # there has to be a faster way...

      return this==that

目标是在进行比较之前确保this和that使用相同的编码。但这似乎是一种尴尬的方式。有更简洁的方法吗？

Answer 1

我认为最简单的方法是将两个输入转换为str并进行比较。

def compare(this, that, encoding="utf-8"):
    // convert this to str
    if isinstance(this, bytes):
        this = str(this, encoding)
    // convert that to str
    if isinstance(that, bytes):
        that = str(that, encoding)
    return this == that

在python中从未知类型转换为UTF-8？

1 个答案: