最近需要写脚本,有个要取汉字拼音首字母的需求,上网查了一些材料,发现很容易实现,提出来大家分享。本脚本用于汉字的拼音的首个字母,如:”我是中国人“,得出的字母为:wszgr。
- #!/usr/bin/env python
- # -*- coding: utf-8 -*-
- def multi_get_letter(str_input):
- if isinstance(str_input, unicode):
- unicode_str = str_input
- else:
- try:
- unicode_str = str_input.decode('utf8')
- except:
- try:
- unicode_str = str_input.decode('gbk')
- except:
- print 'unknown coding'
- return
- return_list = []
- for one_unicode in unicode_str:
- return_list.append(single_get_first(one_unicode))
- return return_list
- def single_get_first(unicode1):
- str1 = unicode1.encode('gbk')
- try:
- ord(str1)
- return str1
- except:
- asc = ord(str1[0]) * 256 + ord(str1[1]) - 65536
- if asc >= -20319 and asc <= -20284:
- return 'a'
- if asc >= -20283 and asc <= -19776:
- return 'b'
- if asc >= -19775 and asc <= -19219:
- return 'c'
- if asc >= -19218 and asc <= -18711:
- return 'd'
- if asc >= -18710 and asc <= -18527:
- return 'e'
- if asc >= -18526 and asc <= -18240:
- return 'f'
- if asc >= -18239 and asc <= -17923:
- return 'g'
- if asc >= -17922 and asc <= -17418:
- return 'h'
- if asc >= -17417 and asc <= -16475:
- return 'j'
- if asc >= -16474 and asc <= -16213:
- return 'k'
- if asc >= -16212 and asc <= -15641:
- return 'l'
- if asc >= -15640 and asc <= -15166:
- return 'm'
- if asc >= -15165 and asc <= -14923:
- return 'n'
- if asc >= -14922 and asc <= -14915:
- return 'o'
- if asc >= -14914 and asc <= -14631:
- return 'p'
- if asc >= -14630 and asc <= -14150:
- return 'q'
- if asc >= -14149 and asc <= -14091:
- return 'r'
- if asc >= -14090 and asc <= -13119:
- return 's'
- if asc >= -13118 and asc <= -12839:
- return 't'
- if asc >= -12838 and asc <= -12557:
- return 'w'
- if asc >= -12556 and asc <= -11848:
- return 'x'
- if asc >= -11847 and asc <= -11056:
- return 'y'
- if asc >= -11055 and asc <= -10247:
- return 'z'
- return ''
- def main(str_input):
- a = multi_get_letter(str_input)
- b = ''
- for i in a:
- b= b+i
- print b
- if __name__ == "__main__":
- str_input='我是中国人'
- main(str_input)
运行如下:
本文转自 lover00751CTO博客,原文链接:http://blog.51cto.com/wangwei007/983289,如需转载请自行联系原作者