闲话少叙，python中经常会对字典的索引进行搜索判断，如判断‘user’是否为{'user1':'New Student','user2':'Old student'}的索引，本文总结了5种方法进行索引的搜索，并比较了运行时间。
四种方法有：
① index in dict.keys()
② dict.has_key(index)
③ index in dict
④ index in set(dict)
⑤ index in set(dict.keys())
先构建一个程序运行时间的函数，用于测试。
from time import clock as now
def costTime(f, testDict, num, describe):
    start = now()
    f(testDict, num)
    finish = now()
    return 'The costing time of %s is %f' % (describe, finish - start)

然后分别写出五个函数：
#测试dict.keys()
def test_dictKeys(testDict, num):
    for i in range(num):
        if i in testDict.keys():
            testDict[i]
           

#测试dict.has_key()
def test_has_key(testDict, num):
    for i in range(num):
        if testDict.has_key(i):
            testDict[i]

#测试dict
def test_dict(testDict, num):           
    for i in range(num):
        if i in testDict:
            testDict[i]

#测试set
def test_set(testDict, num):           
    tt = set(testDict)
    for i in range(num):
        if i in tt:
            testDict[i]

#测试set_dictkeys
def test_setDictKeys(testDict, num):
    tt = set(testDict.keys())
    for i in range(num):
        if i in tt:
            testDict[i]

下面为主程序
num = 50000
#初始化字典
test = dict()
for i in range(num):
    test[i] = i-0.1
   
print costTime(test_dictKeys, test,  num, 'dict.keys()')
print costTime(test_has_key, test, num, 'dict.has_key()')
print costTime(test_dict, test, num, 'in dict')
print costTime(test_set, test, num, 'in set')
print costTime(test_setDictKeys, test, num, 'set(dict.keys())')

得到的结果为：
The costing time of dict.keys() is 79.946342
The costing time of dict.has_key() is 0.017616
The costing time of in dict is 0.012359
The costing time of in set is 0.015817
The costing time of set(dict.keys()) is 0.016689

从以上运行时间上，可以看出第三种方法是最快速的，第一种方法是最慢的，而且相差不是一星半点。究其原因，应该是dict.keys()得到的结果是有序的list数据类型，而最后面三种是dict或者set类型，是无序的。但更深层次的解释，还需要继续深究。也请大神们赐教。