Python timeit 模块，py() 实例源码

def do(setup_statements, statement):
    # extracted from timeit.py
    t = timeit.Timer(stmt=statement,
                     setup="\n".join(setup_statements))
    # determine number so that 1.0 <= total time < 10.0
    for i in range(1, 10):
        number = 10**i
        x = t.timeit(number)
        if x >= 1.0:
            break
    return x / number

def do(setup_statements, statement):
    # extracted from timeit.py
    t = timeit.Timer(stmt=statement,
                     setup="\n".join(setup_statements))
    # determine number so that 0.2 <= total time < 2.0
    for i in range(1, 10):
        number = 10**i
        x = t.timeit(number)
        if x >= 0.5:
            break
    return x / number

def do(setup_statements, statement):
    # extracted from timeit.py
    t = timeit.Timer(stmt=statement,
                     setup="\n".join(setup_statements))
    # determine number so that 0.2 <= total time < 2.0
    for i in range(1, 10):
        number = 10**i
        x = t.timeit(number)
        if x >= 1.0:
            break
    return x / number

def do(setup_statements, statement):
    # extracted from timeit.py
    t = timeit.Timer(stmt=statement,
                     setup="\n".join(setup_statements))
    # determine number so that 1.0 <= total time < 10.0
    for i in range(1, 10):
        number = 10**i
        x = t.timeit(number)
        if x >= 1.0:
            break
    return x / number

def do(setup_statements, statement):
    # extracted from timeit.py
    t = timeit.Timer(stmt=statement,
                     setup="\n".join(setup_statements))
    # determine number so that 0.2 <= total time < 2.0
    for i in range(1, 10):
        number = 10**i
        x = t.timeit(number)
        if x >= 0.5:
            break
    return x / number

def run(self):
        def do(setup_statements, statement):
            # extracted from timeit.py
            t = timeit.Timer(stmt=statement,
                             setup="\n".join(setup_statements))
            # determine number so that 0.2 <= total time < 2.0
            for i in range(1, 10):
                number = 10**i
                x = t.timeit(number)
                if x >= 0.2:
                    break
            return x / number

        for curve in ["NIST192p", "NIST224p", "NIST256p", "SECP256k1",
                      "NIST384p", "NIST521p"]:
            S1 = "from ecdsa import six, SigningKey, %s" % curve
            S2 = "sk = SigningKey.generate(%s)" % curve
            S3 = "msg = six.b('msg')"
            S4 = "sig = sk.sign(msg)"
            S5 = "vk = sk.get_verifying_key()"
            S6 = "vk.verify(sig, msg)"
            # We happen to know that .generate() also calculates the
            # verifying key, which is the time-consuming part. If the code
            # were changed to lazily calculate vk, we'd need to change this
            # benchmark to loop over S5 instead of S2
            keygen = do([S1], S2)
            sign = do([S1,S2,S3], S4)
            verf = do([S1,S2,S3,S4,S5], S6)
            import ecdsa
            c = getattr(ecdsa, curve)
            sig = ecdsa.SigningKey.generate(c).sign(ecdsa.six.b("msg"))
            print("%9s: siglen=%3d, keygen=%.3fs, sign=%.3fs, verify=%.3fs" \
                  % (curve, len(sig), keygen, sign, verf))