如何在不同的编程语言中自然地对字符串数组进行排序？发布您的实现以及答案中的语言.
以下是如何在Python中获得类似资源管理器的行为:
#!/usr/bin/env python
"""
>>> items = u'a1 a003 b2 a2 a10 1 10 20 2 c100'.split()
>>> items.sort(explorer_cmp)
>>> for s in items:
... print s,
1 2 10 20 a1 a2 a003 a10 b2 c100
>>> items.sort(key=natural_key, reverse=True)
>>> for s in items:
... print s,
c100 b2 a10 a003 a2 a1 20 10 2 1
"""
import re
def natural_key(astr):
"""See http://www.codinghorror.com/blog/archives/001018.html"""
return [int(s) if s.isdigit() else s for s in re.split(r'(\d+)', astr)]
def natural_cmp(a, b):
return cmp(natural_key(a), natural_key(b))
try: # use explorer's comparison function if available
import ctypes
explorer_cmp = ctypes.windll.shlwapi.StrCmpLogicalW
except (ImportError, AttributeError):
# not on Windows or old python version
explorer_cmp = natural_cmp
if __name__ == '__main__':
import doctest; doctest.testmod()
要支持Unicode字符串,.isdecimal()应该使用而不是.isdigit().
.isdigit()int()在某些语言环境中,Python 2上的字节串也可能失败(返回值不被接受),例如,Windows上的cp1252语言环境中的'\ xb2'('²').
JavaScript的
Array.prototype.alphanumSort = function(caseInsensitive) {
for (var z = 0, t; t = this[z]; z++) {
this[z] = [], x = 0, y = -1, n = 0, i, j;
while (i = (j = t.charAt(x++)).charCodeAt(0)) {
var m = (i == 46 || (i >=48 && i <= 57));
if (m !== n) {
this[z][++y] = "";
n = m;
}
this[z][y] += j;
}
}
this.sort(function(a, b) {
for (var x = 0, aa, bb; (aa = a[x]) && (bb = b[x]); x++) {
if (caseInsensitive) {
aa = aa.toLowerCase();
bb = bb.toLowerCase();
}
if (aa !== bb) {
var c = Number(aa), d = Number(bb);
if (c == aa && d == bb) {
return c - d;
} else return (aa > bb) ? 1 : -1;
}
}
return a.length - b.length;
});
for (var z = 0; z < this.length; z++)
this[z] = this[z].join("");
}
资源
对于MySQL,我个人使用Drupal模块中的代码,该模块可从hhttp://drupalcode.org/project/natsort.git/blob/refs/heads/5.x-1.x获得:/natsort.install.mysql
基本上,您执行发布的SQL脚本来创建函数,然后使用 ORDER BY natsort_canon(field_name, 'natural')
以下是该函数的自述文件:http://drupalcode.org/project/natsort.git/blob/refs/heads/5.x-1.x: /README.txt
下面是清理代码中的文章链接到的问题:
def sorted_nicely(strings):
"Sort strings the way humans are said to expect."
return sorted(strings, key=natural_sort_key)
def natural_sort_key(key):
import re
return [int(t) if t.isdigit() else t for t in re.split(r'(\d+)', key)]
但实际上我没有机会以这种方式排序.
京公网安备 11010802040832号 | 京ICP备19059560号-6 