Rのpower.prop.test関数をpythonで実装する #Python

適合度テストを設計するのに、サンプルサイズや検証力を求めたい
Rのpower.prop.testという関数を使えば求められる
同じような関数をpythonでも使いたい…

ということで、練習としてRのpower.prop.test関数のソースコードを元にpythonで実装してみました。
引数の意味などはRの公式ドキュメントを参照してください。
コードは記事の最後に載せてます。

使用例

サンプルサイズを求める

グループ1の比率は1%、グループ2の比率は3%
有意水準は0.05
検出力は0.8
片側検定

上記の条件を満たすためのサンプルの大きさnを求めます。

# python
> params_dict = power_prop_test(p1=0.01, p2=0.03, power=0.8, sig_level=0.05, alternative="one_sided")

    Sample size: n = 604.845426012357
    Probability: p1 = 0.01, p2 = 0.03
    sig_level = 0.05,
    power = 0.8,
    alternative = one_sided

    NOTE: n is number in "each" group

> print(params_dict)

{'n': 604.845426012357, 'p1': 0.01, 'p2': 0.03, 'sig_level': 0.05, 'power': 0.8, 'alternative': 'one_sided'}

Rで同じ値を渡した結果とほぼ一致します。

# R
> power.prop.test(p1=0.01, p2=0.03, power=0.8, sig.level=0.05, alternative="one.sided")

     Two-sample comparison of proportions power calculation 

              n = 604.8434
             p1 = 0.01
             p2 = 0.03
      sig.level = 0.05
          power = 0.8
    alternative = one.sided

NOTE: n is number in *each* group

検出力を求める

次に、サンプルサイズn=1000のときの検出力(power)について同様に求めてみます。

#python
> power_prop_test(n=1000, p1=0.01, p2=0.03, sig_level=0.05, alternative="one_sided")

    Sample size: n = 1000
    Probability: p1 = 0.01, p2 = 0.03
    sig_level = 0.05,
    power = 0.9398478094069961,
    alternative = one_sided

    NOTE: n is number in "each" group

#R
> power.prop.test(n=1000, p1=0.01, p2=0.03, sig.level=0.05, alternative="one.sided")

     Two-sample comparison of proportions power calculation 

              n = 1000
             p1 = 0.01
             p2 = 0.03
      sig.level = 0.05
          power = 0.9398478
    alternative = one.sided

NOTE: n is number in *each* group

こちらもほとんど同じ値をとります。

補足

python上でRを動かすPypeRなどで、直接Rの関数を呼び出したほうが簡単かも
初期値によっては収束せず、エラーを吐くかもしれません

読んでいただきありがとうございました。

使用したpower_prop_test関数

# 求めたいパラメーターをNoneにする
# alternativeは"one_sided" "two_sided"のどちらかを選択
# 求めた各パラメータを辞書型で返します
def power_prop_test(n=None, p1=None, p2=None, sig_level=0.05, power=None,
                    alternative="one_sided", strict=False,
                    tol=2.220446049250313e-16**0.25):

    from math import sqrt
    from scipy.stats import norm
    from scipy.optimize import brentq

    missing_params = [arg for arg in [n, p1, p2, sig_level, power] if not arg]
    num_none = len(missing_params)

    if num_none != 1:
        raise ValueError("exactly one of 'n', 'p1', 'p2', 'power' and 'sig.level'must be None")

    if sig_level is not None:
        if sig_level < 0 or sig_level > 1:
            raise ValueError("\"sig_level\" must be between 0 and 1")

    if power is not None:
        if power < 0 or power > 1:
            raise ValueError("\"power\" must be between 0 and 1")

    if alternative not in ["two_sided", "one_sided"]:
        raise ValueError("alternative must be \"two_sided\" or \"one_sided\"")
    t_side_dict = {"two_sided":2, "one_sided":1}
    t_side = t_side_dict[alternative]

    # compute power 
    def p_body(p1=p1, p2=p2, sig_level=sig_level, n=n, t_side=t_side, strict=strict):
        if strict and t_side==2:
            qu = norm.ppf(1-sig_level/t_side)
            d = abs(p1-p2)
            q1 = 1-p1
            q2 = 1-p2
            pbar = (p1 + p2) / 2
            qbar = 1 - pbar
            v1 = p1 * q1
            v2 = p2 * q2
            vbar = pbar * qbar
            power_value = (norm.cdf((sqrt(n) * d - qu * sqrt(2 * vbar))/sqrt(v1 + v2))
                    + norm.cdf(-(sqrt(n) * d + qu * sqrt(2 * vbar))/sqrt(v1 + v2)))
            return power_value

        else:
            power_value = norm.cdf((sqrt(n) * abs(p1 - p2) - (-norm.ppf(sig_level / t_side)
                        * sqrt((p1 + p2) * (1 - (p1 + p2)/2)))) / sqrt(p1 * 
                        (1 - p1) + p2 * (1 - p2)))
            return power_value

    if power is None:
        power = p_body()

    # solve the equation for the None value argument 
    elif n is None:
        def n_solver(x, power=power):
            return p_body(n=x) - power
        n = brentq(f=n_solver, a=1, b=1e+07, rtol=tol, maxiter=1000000)

    elif p1 is None:
        def p1_solver(x, power=power):
            return p_body(p1=x) - power
        try:
            p1 = brentq(f=p1_solver, a=0, b=p2, rtol=tol, maxiter=1000000)
        except:
            ValueError("No p1 in [0, p2] can be found to achive the desired power")

    elif p2 is None:
        def p2_solver(x, power=power):
            return p_body(p2=x) - power
        try:
            p2 = brentq(f=p2_solver, a=p1, b=1, rtol=tol, maxiter=1000000)
        except:
            VealueError("No p2 in [p1, 1] can be found to achive the desired power")

    elif sig_level is None:
        def sig_level_solver(x, power=power):
            return p_body(sig_level=x) - power
        try:
            sig_level = brentq(f=sig_level_solver, a=1e-10, b=1-1e-10, rtol=tol, maxiter=1000000)
        except:
            ValueError("No significance level [0,1] can be found to achieve the desired power")

    print("""
    Sample size: n = {0}
    Probability: p1 = {1}, p2 = {2}
    sig_level = {3},
    power = {4},
    alternative = {5}

    NOTE: n is number in "each" group
    """.format(n, p1, p2, sig_level, power, alternative))
    #各パラメーターは引数名をキーにもつ辞書で返される
    params_dict = {"n":n, "p1":p1, "p2":p2,
                   "sig_level":sig_level, "power":power, "alternative":alternative}

    return params_dict