假设以下数据：

library(cobalt)
library(dplyr)

set.seed(123)

# Extending lalonde data by `group` variable:
lalonde <- cbind(lalonde,
                 group = sample(c("A", "B", "AB"), size = 614, replace = TRUE))

# Creating variable `set` for AB group:
lalonde$set[lalonde$group == "AB"] <- sample(c(5, 10, 15, 3), sum(lalonde$group == "AB"), replace = TRUE)

# Filling 'set' column for `group` 'A' and 'B':
prob_80 <- sample(c(0, 1), nrow(lalonde), replace = TRUE, prob = c(0.8, 0.2))
lalonde$set[(lalonde$group == "A" | lalonde$group == "B") & prob_80 == 1] <- sample(c(5, 10, 15, 3), sum((lalonde$group == "A" | lalonde$group == "B") & prob_80 == 1), replace = TRUE)
lalonde$set[(lalonde$group == "A" | lalonde$group == "B") & is.na(lalonde$set)] <- sample(1:100, sum((lalonde$group == "A" | lalonde$group == "B") & is.na(lalonde$set)), replace = TRUE)

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现在我们有一个group变量，包含A、B、AB和一个名为set的变量。
现在我想拟合一个逻辑PS模型，该模型由group == "AB"的set值分层，预测在AB组中。
首先，我将提取group == AB中set的不同值。

unique_set_values <- unique(lalonde[lalonde$group == "AB", "set"])  %>% 
  print()

型
它们是：

+   print()
[1]  5 15 10  3

型
我使用它们来获取属于set值之一的所有观测值：

filtered_data <- lalonde %>% 
  filter(set %in% unique_set_values)

型
然后我将数据拆分，并将AB替换为1，否则为0：

# For AB and A:
AB_A <- filtered_data %>% 
  filter(group %in% c("AB", "A")) %>%
  mutate(group = ifelse(group == "AB", 1, 0)) 

# For AB and B:
AB_B <- filtered_data %>% 
  filter(group %in% c("AB", "B")) %>%
  mutate(group = ifelse(group == "AB", 1, 0))

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现在我可以计算AB和A以及AB和B的分层PS：

# Creating a formula:
formula <- group ~ age + educ + race + married + nodegree + re74 + re75 + re78

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但是在这种情况下，如何计算set分层的PS？
我试过了：

AB_A_PS <- AB_A %>%
  group_by(set) %>%
  mutate(pscore = glm(formula, data = ., family = binomial(link = "logit"))$fitted.values)

型
但我得到的是一个错误：

Error in `mutate()`:
ℹ In argument: `pscore = predict(glm(formula, data = ., family = binomial(link = "logit")))`.
ℹ In group 1: `set = 3`.
Caused by error:
! `pscore` must be size 66 or 1, not 238.

所以，很明显，它不起作用。
谢谢你