Groupon 案例
Groupon 是一家致力於利用社交媒體協調團購的電子商務公司。為了取得購物折扣，Groupon團購會要求達到某種條件的最低要求，買家才能享有折扣，例如，最低購買人數。這類的最低購買條件是否會影響交易結果呢？Propensity score matching (PSM) 可以幫助我們用統計的方法回答此類的問題。以下我們定義一個二元自變數(Independent Varaible, IV)
- 對照組: 無最低購買人數要求的優惠券(0)
- 處理組: 有最低購買人數要求的優惠券(1)
範例資料 groupon.csv 有下示結構，其中包含三個依變數：收入 、售出數量和臉書讚數:
# <class 'pandas.core.frame.DataFrame'>
# RangeIndex: 710 entries, 0 to 709
# Data columns (total 13 columns):
# # Column Non-Null Count Dtype
# --- ------ -------------- -----
# 0 deal_id 710 non-null object
# 1 start_date 710 non-null object
# 2 min_req 710 non-null int64
# 3 treatment 710 non-null int64
# 4 prom_length 710 non-null int64
# 5 price 710 non-null int64
# 6 discount_pct 710 non-null int64
# 7 coupon_duration 710 non-null int64
# 8 featured 710 non-null int64
# 9 limited_supply 710 non-null int64
# 10 fb_likes 710 non-null int64
# 11 quantity_sold 710 non-null int64
# 12 revenue 710 non-null int64
# dtypes: int64(11), object(2)
# memory usage: 72.2+ KB
df = pd.read_csv('data/groupon.csv')
df.info()
grouped = df.groupby("treatment")
grouped[["fb_likes", "quantity_sold", "revenue"]].mean()
# fb_likes quantity_sold revenue
# treatment
# 0 77.941296 333.002024 9720.987854
# 1 113.203704 509.351852 12750.694444
在開始分析前，我們先引入多項工具包，並自定二個函數。
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import seaborn as sns
import math
from scipy.stats import ttest_ind
from sklearn.linear_model import LogisticRegression
from sklearn.neighbors import NearestNeighbors
def t_test(df_control, df_treatment, DV="revenue", alpha=0.05):
ctl_avg = df_control[DV].mean()
trt_avg = df_treatment[DV].mean()
# compare samples
_, p = ttest_ind(df_control[DV], df_treatment[DV])
print(f"p = {p:.3f}")
print(f"Control vs. treatment: {ctl_avg:.2f} vs. {trt_avg:.2f}")
# interpret
if p > alpha:
print(
"H0 is plausible."
)
else:
print("The group means differ")
return p
def logit(p):
# The logit function is defined as the logarithm of the odds of the probability p of a certain event occurring:
# Logit(p) = log(p / (1 - p))
return math.log(p / (1 - p))在未使用傾向評分進行配對前,t-test顯示，當優恵卷附加最低購買需求時，似乎和高收益、更多臉書讚、及售出更多商品是相關的。
# student's t-test for revenue (dependent variable)
# separate control and treatment for t-test
df_control = df[df.treatment == 0]
df_treatment = df[df.treatment == 1]
p_revenue = t_test(df_control, df_treatment, DV="revenue")
p_fb = t_test(df_control, df_treatment, DV="fb_likes")
p_qs = t_test(df_control, df_treatment, DV="quantity_sold")
# p = 0.040
# Control vs. treatment: 9720.99 vs. 12750.69
# Different group means (reject H0)
# p = 0.004
# Control vs. treatment: 77.94 vs. 113.20
# Different group means (reject H0)
# p = 0.001
# Control vs. treatment: 333.00 vs. 509.35
# Different group means (reject H0)
多元邏輯回歸模型 （Multiple Logistic Regression）
當我們使用邏輯回歸將多個共變數轉換為機率值，此概率便是傾向評分 （PS），它表示觀察結果屬於處理組的概率（1 = 處理，0 = 對照）。
基本上，這是一個分類問題：根據樣本的特徵，來預測樣本屬於處理組中的概率。在進行邏輯回歸前，我們必須先審慎地選擇適合的變數。以下兩項原則可幫助排除一些不適當的變數：
- 和自變項相關係數接近1的變數
- 和自變項有高相關的變數
在此分析中，我們選擇了六個變數：prom_length、price、discount_pct、coupon_duration、featured 和 limited_supply來訓練邏輯回歸模型。在邏輯回歸模型中，目標變數--屬於處理理或是對照組--是被視為依變數，而不是如同我們的主要研究問題將之視為自變數。
X = df[['prom_length', 'price', '適discount_pct', 'coupon_duration', 'featured','limited_supply']]
y = df['treatment']
lr = LogisticRegression()
lr.fit(X, y)
pred_prob = lr.predict_proba(X) # probabilities for classes
# the propensity score (ps) is the probability of being 1 (i.e., in the treatment group)
df['ps'] = pred_prob[:, 1]
sns.histplot(data=df, x='ps', hue='treatment')
以下直方圖顯示對照組和處理組在傾向評分上的有相當程序的重合，因此我們應該可以找到適當數量的配對。
KNN模型
PSM的演算法類似於處理單一共變數的情況。第一、我們將每一個處理組中的樣本依序取出，第二、在對照組中找出與此處理組樣本在傾向分數上相近的樣本。在此我們必須了解傾向分數是連續的數值資料，所以兩個數字是不可能100%完全相同，因此PSM演算法的第三步是，使用 KNN 的統計模型計算每一樣本之間的相對傾向分數（稱之為距離），第四、據此距離參數找出每一個樣本以及與它鄰近的其他 N ，例如，9個樣本。
- 在開始之前，我們必須先定義一鄰近半徑以來決定誰是鄰居誰不是有鄰居，例如我們可以使用25%的傾向分數標準差做為半徑。
- 當使用一對多匹配 （此例我們使用一對九），我們從前 N 個鄰居中選擇 M 個最接近的匹配項 (此例M=1)。
因為半徑定義未考量所有的傾向分數，所以有些處理組的樣本可能找不到任何鄰近樣本，這些處理組的樣本將不會被使用到。
# Define 25% of standard deviation of the propensity score as the caliper
caliper = np.std(df.ps) * 0.25
print(f"caliper (radius) is: {caliper:.4f}")
# Get the 10 closest neighbors
n_neighbors = 10
# setup knn
knn = NearestNeighbors(n_neighbors=n_neighbors, radius=caliper)
ps = df[["ps"]]
knn.fit(ps)
下面的程式碼檢驗KNN模型所儲存的距離及序號參數，此例中 K=9 ，因此distance[0]的第一個元素是參考點，其距離值被定為0，其他樣本的距離值則是相對於此參考點，distance儲存的其他九個元素則列出距此參考點最近的九個資料點。
# distances and indexes
distances, neighbor_indexes = knn.kneighbors(ps)
# the 10 closest points to the first point
print(distances[0])
# [0.00000000e+00 7.98366770e-05 3.78158885e-04 7.53425375e-04
# 1.18941498e-03 1.37592834e-03 1.37592834e-03 1.62491413e-03
# 1.67879150e-03 2.01192411e-03]
print(neighbor_indexes.shape)
# (710, 10)
# [0 348 388 415 624 494 150 463 372 345]
neighbor_indexes儲存樣本於原來資料檔中的序號。
df.loc[neighbor_indexes[0], ["prom_length", "price", "treatment", "ps"]]
# prom_length price treatment ps
# 0 4 99 1 0.259179
# 348 3 19 0 0.259099
# 388 3 20 0 0.258801
# 415 3 15 0 0.258425
# 624 3 50 0 0.260368
# 494 3 29 0 0.260555
# 150 3 29 1 0.260555
# 463 3 30 0 0.257554
# 372 4 16 0 0.260858
# 345 3 25 0 0.257167
matched_control = [] # a container to store the matched sample
# 1. iterate over the dataframe
for current_index, row in df.iterrows():
# 2. When the current row is in the control group,
if row.treatment == 0:
# set matched column to nan
df.loc[current_index, "matched"] = np.nan
else:
for idx in neighbor_indexes[
current_index, :
]: # for each row in treatment, find the k neighbors
# 3. make sure the current row is not itself
# 4. and the neighbor is in the control
if (current_index != idx) and (df.loc[idx].treatment == 0):
# 5. sampling without replacement - if this control sample
# has not been matched yet,
if idx not in matched_control:
# we record the index of the matching sample and
df.loc[current_index, "matched"] = idx
# then append it to the matched list
matched_control.append(idx)
break
df.treatment.value_counts()
# treatment
# 0 494
# 1 216
# Name: count, dtype: int64
# dfmatched.value_counts()
s = df[["matched"]]
nan_count = s.isna().sum() # or s.isnull().sum() for older pandas versions
nan_count
# matched 526
# dtype: int64
在取出匹配後的處理組及對照組，我們將之重新整理為一個新的資料集。
# control have no match
treatment_matched = df.dropna(subset=["matched"]) # drop not matched
# matched control observation indexes
control_matched_idx = treatment_matched.matched
control_matched_idx = control_matched_idx.astype(int) # change to int
control_matched = df.loc[control_matched_idx, :] # select matched control observations
# combine the matched treatment and control
df_matched = pd.concat([treatment_matched, control_matched])
df_matched.treatment.value_counts()最後，我們使用Cohen's d的效果值來檢驗匹配前後的異同。
from numpy import mean
from numpy import var
from math import sqrt
# function to calculate Cohen's d for independent samples
def cohen_d(d1, d2):
# calculate the size of samples
n1, n2 = len(d1), len(d2)
# calculate the variance of the samples
s1, s2 = var(d1, ddof=1), var(d2, ddof=1)
# calculate the pooled standard deviation
s = sqrt(((n1 - 1) * s1 + (n2 - 1) * s2) / (n1 + n2 - 2))
# calculate the means of the samples
u1, u2 = mean(d1), mean(d2)
# calculate the effect size
return (u1 - u2) / s
effect_sizes = []
cols = [
"revenue",
"fb_likes",
"quantity_sold",
"prom_length",
"price",
"discount_pct",
"coupon_duration",
"featured",
"limited_supply",
]
for cl in cols:
p_before = t_test(df_control, df_treatment, DV = cl)
p_after = t_test(df_matched_control, df_matched_treatment, DV = cl)
cohen_d_before = cohen_d(df_treatment[cl], df_control[cl])
cohen_d_after = cohen_d(df_matched_treatment[cl], df_matched_control[cl])
effect_sizes.append([cl, "before", cohen_d_before, p_before])
effect_sizes.append([cl, "after", cohen_d_after, p_after])
df_effect_sizes = pd.DataFrame(
effect_sizes, columns=["feature", "matching", "effect_size", "p-value"]
)
fig, ax = plt.subplots(figsize=(15, 5))
bar_plot = sns.barplot(
data=df_effect_sizes, x="effect_size", y="feature", hue="matching", orient="h"
)
fig = bar_plot.get_figure()
fig.savefig("horizontal_bar.png")
傾向評分匹配是一種包含多個步驟的共變數分析，它的目的是在於平衡多個共變數對在二元自變項的影響，進而得到更接近且反映自然現象的結果。然而，在此例中我們也看到了，使用PSM可能會有許多處理組樣本找不到適當的對照組匹配，因此若這些樣本確有價值，而我們沒有使用到，導致部份資料流失。無論如何，在此例中，使用或是不使用PSM都沒有改變三項依變須的統計檢驗結果。歸根結底，是否該使用PSM仍須依靠研究者根據個別情況做分項考量。。
參考資料
- groupon.csv 來自於 Gang Wang's kaggle page.