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Shapley Additive Explanations（SHAP 值）的效用是了解每个特征如何对模型的预测做出贡献。对于某些目标，例如以 RMSE 作为目标函数的回归，SHAP 值采用标签值的本机单位。例如，如果估算住房成本，SHAP 值可以表示为美元。正如您将在下面看到的，并非所有目标函数都是如此。特别是，Tweedie 回归目标不会产生原生单位的 SHAP 值。这是一个解释问题，因为我们想知道住房成本如何受到 +/- 美元特征的影响。

鉴于这些信息，我的问题是：在解释具有 Tweedie 回归目标的模型时，我们如何将每个单独特征的 SHAP 值转换为目标标签的数据空间？

我不知道目前有任何包实现了这种转换。这在 shap 作者自己发布的包中仍未解决。

我在下面用 lightgbm 的 R 实现来说明这个问题的细节：

library(tweedie)
library(lightgbm)

set.seed(123)

tweedie_variance_power <- 1.2

labels <- rtweedie(1000, mu = 1, phi = 1, power = tweedie_variance_power)
hist(labels)

feat1 <- labels + rnorm(1000) #good signal for label with some noise
feat2 <-rnorm(1000) #garbage feature 
feat3 <-rnorm(1000) #garbage feature 

features <- cbind(feat1, feat2, feat3)

dTrain <- lgb.Dataset(data = features,
                      label = labels)

params <- c(objective = 'tweedie',
            tweedie_variance_power = tweedie_variance_power)

mod <- lgb.train(data = dTrain,
                 params = params,
                 nrounds = 100)

#Predictions in the native units of the labels
predsNative <- predict(mod, features, rawscore = FALSE)
#Predictions in the raw format
predsRaw <- predict(mod, features, rawscore = TRUE)

#We do not expect these values to be equal
all.equal(predsTrans, predsRaw)
"Mean relative difference: 1.503072"

#We expect values to be equal if raw scores are exponentiated
all.equal(predsTrans, exp(predsRaw))
"TRUE" #... our expectations are correct

#SHAP values 
shapNative <- predict(mod, features, rawscore = FALSE, predcontrib = TRUE)
shapRaw <- predict(mod, features, rawscore = TRUE, predcontrib = TRUE )

#Are there differences between shap values when rawscore is TRUE or FALSE?
all.equal(shapNative, shapRaw)
"TRUE" #outputs are identical, that is surprising!

#So are the shap values in raw or native formats?
#To anwser this question we can sum them

#testing raw the raw case first
all.equal(rowSums(shapRaw), predsRaw)
"TRUE" 

#from this we can conclude that shap values are not in native units,
#regardless of whether rawscore is TRUE or FALSE

#Test native scores just to prove point
all.equal(rowSums(shapNative), predsNative)
"Mean relative difference: 1.636892" # reaffirms that shap values are not in native units

#However, we can perform this operation on the raw shap scores
#to get the prediction in the native value
all.equal(exp(rowSums(shapRaw)), predsNative)
'TRUE'

#reversing the operations does not yield the same result
all.equal(rowSums(exp(shapRaw)), predsNative)
"Mean relative difference: 0.7662481"

#The last line is relevant because it implies 
#The relationship between native predictions
#and exponentiated shap values is not linear

#So, given the point of SHAP is to understand how each 
#feature impacts the prediction in its native units
#the raw shap values are not as useful as they could be

#Thus, how how would we convert 
#each of these four raw shap value elements to native units,
#thus understanding their contributions to their predictions
#in currency of native units?
shapRaw[1,]
-0.15429227  0.04858757 -0.27715359 -0.48454457

原始帖子和编辑

我对 SHAP 值的理解是，它们在进行回归时采用标签/响应的本机单位，并且 SHAP 值的总和近似于模型的预测。

我正在尝试使用 Tweedie 回归目标提取 LightGBM 包中的 SHAP 值，但发现 SHAP 值不在标签的本机单位中，并且它们不与预测值相加。

看来它们必须取幂，这是正确的吗？

旁注：我了解 SHAP 值矩阵的最后一列代表基本预测，必须添加。

可重现的例子：

library(tweedie)
library(caret)
library(lightgbm)

set.seed(123)

tweedie_variance_power <- 1.2

labels <- rtweedie(1000, mu = 1, phi = 1, power = tweedie_variance_power)
hist(labels)

feat1 <- labels + rnorm(1000) #good signal for label with some noise
feat2 <-rnorm(1000) #garbage feature 
feat3 <-rnorm(1000) #garbage feature 

features <- cbind(feat1, feat2, feat3)

dTrain <- lgb.Dataset(data = features,
                      label = labels)

params <- c(objective = 'tweedie',
            tweedie_variance_power = tweedie_variance_power)

mod <- lgb.train(data = dTrain,
                 params = params,
                 nrounds = 100)

preds <- predict(mod, features)

plot(preds, labels,
     main = paste('RMSE =', 
                  RMSE(pred = preds, obs = labels)))

#shap values are summing to negative values?
shap_vals <- predict(mod, features, predcontrib = TRUE, rawscore = FALSE)
shaps_sum <- rowSums(shap_vals)
plot(shaps_sum, labels, 
     main = paste('RMSE =', 
                  RMSE(pred = shaps_sum, obs = labels)))

#maybe we need to exponentiate?
shap_vals_exp <- exp(shap_vals)
shap_vals_exp_sum <- rowSums(shap_vals_exp)
#still looks a little weird, overpredicting 
plot(shap_vals_exp_sum, labels,
     main = paste('RMSE =',
                  RMSE(pred = shap_vals_exp_sum, obs = labels)))

编辑

操作的顺序是先求和，然后对 SHAP 值求幂，这将为您提供本机单位的预测。虽然我仍然不清楚如何将特征级别值转换为原生响应单元。

shap_vals_sum_exp <- exp(shaps_sum)
plot(shap_vals_sum_exp, labels,
     main = paste('RMSE =',
                  RMSE(pred = shap_vals_sum_exp, obs = labels)))
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1 回答 1

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我将展示如何在 Python 中协调 shap 值和模型预测，包括原始分数和原始单位。希望它能帮助您了解您在 R 中的位置。

步骤 1. 生成数据集

# pip install tweedie
import tweedie
y = tweedie.tweedie(1.2,1,1).rvs(size=1000)
X = np.random.randn(1000,3)

步骤 2. 拟合模型

from lightgbm.sklearn import LGBMRegressor
lgb = LGBMRegressor(objective = 'tweedie')
lgb.fit(X,y)

步骤 3. 了解什么是 shap 值。

第 0 个数据点的形状值

shap_values = lgb.predict(X, pred_contrib=True)
shap_values[0]
array([ 0.36841812, -0.15985678,  0.28910617, -0.27317984])

前 3 个是模型对基线的贡献，即 shap 值本身：

shap_values[0,:3].sum()
0.4976675073764354

第 4 个是原始分数的基线：

shap_values[0,3]
-0.2731798364061747

它们的总和加起来就是原始分数中的模型预测：

shap_values[0,:3].sum() + shap_values[0,3]
0.22448767097026068

让我们检查一下原始模型预测：

preds = lgb.predict(X, raw_score=True)
preds[0]
0.2244876709702609

编辑。原始分数和原始单位之间的转换

要在 Tweedie（以及 Poisson 和 Gamma）分布的原始分数和原始单位之间进行转换，您需要了解 2 个事实：

  1. 原件是exp生的
  2. expsum是product的_exps

演示：

  1. 原始单位的第 0 个预测：
lgb.predict([X[0,:]])
array([0.39394102])
  1. 原始分数空间中第 0 行的形状值：
shap_values = lgb.predict(X, pred_contrib=True, raw_score=True)
shap_values[0]
array([-0.77194274, -0.08343294,  0.22740536, -0.30358374])
  1. 将形状值转换为原始单位（指数乘积）：
np.prod(np.exp(shap_values[0]))
0.3939410249402226

又和我长得一模一样。

于 2020-11-12T08:07:26.930 回答