Python实现检测字符串是否全为汉字（含生僻字）

Python实现检测字符串是否全为汉字（含生僻字） 1.  中文汉字Unicode 编码表

序号字符集字数Unicode 编码范围1基本汉字209024E00-9FA52基本汉字补充749FA6-9FEF3扩展A65823400-4DB54扩展B4271120000-2A6D65扩展C41492A700-2B7346扩展D2222B740-2B81D7扩展E57622B820-2CEA18扩展F74732CEB0-2EBE09康熙部首2142F00-2FD510部首扩展1152E80-2EF311兼容汉字477F900-FAD912兼容扩展5422F800-2FA1D13PUA(GBK)部件81E815-E86F14部件扩展452E400-E5E815PUA增补207E600-E6CF16汉字笔画3631C0-31E317汉字结构122FF0-2FFB18汉语注音433105-312F19注音扩展2231A0-31BA20〇13007 2. Python代码实现

#只要是检测到一个非汉字字符就返回
#if条件一大堆，肯定有更简单的写法，再学吧！
def is_ch(word):
    for ch in word:
        if not('u4e00' <= ch <= 'u9fef') and not ('u3400' <= ch <= 'u4db5') 
                and not ('u20000' <= ch <= 'u2a6d6') and not ('u2a700' <= ch <= 'u2b734')
                and not ('u2b740' <= ch <= 'u2b81d') and not ('u2b820' <= ch <= 'u2cea1')
                and not ('u2ceb0' <= ch <= 'u2ebe0') and not ('u2f00' <= ch <= 'u2fd5')
                and not ('u2e80' <= ch <= 'u2ef3') and not ('uf900' <= ch <= 'ufad9')
                and not ('u2f800' <= ch <= 'u2fa1d') and not ('ue815' <= ch <= 'ue86f')
                and not ('ue400' <= ch <= 'ue5e8') and not ('ue600' <= ch <= 'ue6cf')
                and not ('u31c0' <= ch <= 'u31e3') and not ('u2ff0' <= ch <= 'u2ffb')
                and not ('u3105' <= ch <= 'u312f') and not ('u31a0' <= ch <= 'u31ba'):
            return False
            break
    return True

3. 有时间时可以扩展

   （1）比如：全部为汉字时返回True和原字符串，有非汉字时返回False和非汉字字符串。

   （2）if中判断条件一大堆，肯定有简单的写法，找到一个简单的写法或是优雅点的写法。

   （3）更简单的实现方法？这些Unicode 编码连续吗？找时间研究一下！