アルファベットの文字の対応する位置を取得するための Python のいくつかの方法とパフォーマンスの比較

In [137]: "ABC".index('B')
Out[137]: 1In [138]: "ABC".index('B')+1Out[138]:
 2#或者在前面填充一个字符,这样index就直接得到字母序号：
 In [139]: "_ABC".index("B")
Out[139]: 2

In [140]: ord('B')-64
Out[140]: 2

az = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"_az = "_ABCDEFGHIJKLMNOPQRSTUVWXYZ"azlist = list(az)

azdict = dict(zip(az,range(1,27)))

text = az*1000000 #这个是测试数据#str.find和str.index的是一样的。这里就没必要写了。def azindexstr(text):
    for r in text:
        az.index(r)+1
        passdef _azindexstr(text):
    for r in text:
        _az.index(r)        passdef azindexlist(text):
    for r in text:
        azlist.index(r)        passdef azindexdict(text):
    for r in text:
        azdict.get(r)        passdef azindexdict2(text):
    for r in text:
        azdict[r]        passdef azord(text):
    for r in text:
        ord(r)-64
        passdef azand64(text):
    for r in text:
        ord(r)%64
        pass

In [147]: %timeit azindexstr(text)
1 loop, best of 3: 9.09 s per loop

In [148]: %timeit _azindexstr(text)
1 loop, best of 3: 8.1 s per loop

In [149]: %timeit azindexlist(text)
1 loop, best of 3: 17.1 s per loop

In [150]: %timeit azindexdict(text)
1 loop, best of 3: 4.54 s per loop

In [151]: %timeit azindexdict2(text)
1 loop, best of 3: 1.99 s per loop

In [152]: %timeit azord(text)
1 loop, best of 3: 2.94 s per loop

In [153]: %timeit azand64(text)
1 loop, best of 3: 4.56 s per loop