本文实例总结了Python实现list反转的方法。分享给大家供大家参考。具体实现方法如下:
下面有几个不同实现的函数
import math def resv(li): new = [] if li: cnt = len(li) for i in range(cnt): new.append(li[cnt-i-1]) return new def resv2(li): li.reverse() return li def resv3(li): hcnt = int(math.floor(len(li)/2)) tmp = 0 for i in range(hcnt): tmp = li[i] li[i] = li[-(i+1)] li[-(i+1)] = tmp return li li = [1, 2, 3, 4, 5] print resv(li)
>>> li = ["a", "b", "mpilgrim", "z", "example"] >>> li ['a', 'b', 'mpilgrim', 'z', 'example'] >>> li[0] 'a' >>> li[4] 'example'
>>> li ['a', 'b', 'mpilgrim', 'z', 'example'] >>> li[-1] 'example' >>> li[-3] 'mpilgrim' >>> li ['a', 'b', 'mpilgrim', 'z', 'example'] >>> li[1:3] ['b', 'mpilgrim'] >>> li[1:-1] ['b', 'mpilgrim', 'z'] >>> li[0:3] ['a', 'b', 'mpilgrim']
>>> li ['a', 'b', 'mpilgrim', 'z', 'example'] >>> li.append("new") >>> li ['a', 'b', 'mpilgrim', 'z', 'example', 'new'] >>> li.insert(2, "new") >>> li ['a', 'b', 'new', 'mpilgrim', 'z', 'example', 'new'] >>> li.extend(["two", "elements"]) >>> li ['a', 'b', 'new', 'mpilgrim', 'z', 'example', 'new', 'two', 'elements']
>>> li ['a', 'b', 'new', 'mpilgrim', 'z', 'example', 'new', 'two', 'elements'] >>> li.index("example") 5 >>> li.index("new") 2 >>> li.index("c") Traceback (innermost last): File "<interactive input>", line 1, in ? ValueError: list.index(x): x not in list >>> "c" in li False
>>> li ['a', 'b', 'new', 'mpilgrim', 'z', 'example', 'new', 'two', 'elements'] >>> li.remove("z") >>> li ['a', 'b', 'new', 'mpilgrim', 'example', 'new', 'two', 'elements'] >>> li.remove("new") >>> li ['a', 'b', 'mpilgrim', 'example', 'new', 'two', 'elements'] >>> li.remove("c") Traceback (innermost last): File "<interactive input>", line 1, in ? ValueError: list.remove(x): x not in list >>> li.pop() 'elements' >>> li ['a', 'b', 'mpilgrim', 'example', 'new', 'two']
6.list 运算符
>>> li = ['a', 'b', 'mpilgrim'] >>> li = li + ['example', 'new'] >>> li ['a', 'b', 'mpilgrim', 'example', 'new'] >>> li += ['two'] >>> li ['a', 'b', 'mpilgrim', 'example', 'new', 'two'] >>> li = [1, 2] * 3 >>> li [1, 2, 1, 2, 1, 2]
>>> params = {"server":"mpilgrim", "database":"master", "uid":"sa", "pwd":"secret"} >>> ["%s=%s" % (k, v) for k, v in params.items()] ['server=mpilgrim', 'uid=sa', 'database=master', 'pwd=secret'] >>> ";".join(["%s=%s" % (k, v) for k, v in params.items()]) 'server=mpilgrim;uid=sa;database=master;pwd=secret'
8.分割字符串
>>> li = ['server=mpilgrim', 'uid=sa', 'database=master', 'pwd=secret'] >>> s = ";".join(li) >>> s 'server=mpilgrim;uid=sa;database=master;pwd=secret' >>> s.split(";") ['server=mpilgrim', 'uid=sa', 'database=master', 'pwd=secret'] >>> s.split(";", 1) ['server=mpilgrim', 'uid=sa;database=master;pwd=secret']
9.list的映射解析
>>> li = [1, 9, 8, 4] >>> [elem*2 for elem in li] [2, 18, 16, 8] >>> li [1, 9, 8, 4] >>> li = [elem*2 for elem in li] >>> li [2, 18, 16, 8]
>>> params = {"server":"mpilgrim", "database":"master", "uid":"sa", "pwd":"secret"} >>> params.keys() ['server', 'uid', 'database', 'pwd'] >>> params.values() ['mpilgrim', 'sa', 'master', 'secret'] >>> params.items() [('server', 'mpilgrim'), ('uid', 'sa'), ('database', 'master'), ('pwd', 'secret')] >>> [k for k, v in params.items()] ['server', 'uid', 'database', 'pwd'] >>> [v for k, v in params.items()] ['mpilgrim', 'sa', 'master', 'secret'] >>> ["%s=%s" % (k, v) for k, v in params.items()] ['server=mpilgrim', 'uid=sa', 'database=master', 'pwd=secret']
>>> li = ["a", "mpilgrim", "foo", "b", "c", "b", "d", "d"] >>> [elem for elem in li if len(elem) > 1] ['mpilgrim', 'foo'] >>> [elem for elem in li if elem != "b"] ['a', 'mpilgrim', 'foo', 'c', 'd', 'd'] >>> [elem for elem in li if li.count(elem) == 1] ['a', 'mpilgrim', 'foo', 'c']
希望本文所述对大家的Python程序设计有所帮助。