pairwise.prop.test 在具有多个类别的 R 中

Question

我有以下数据集，其中包含 19 家公司按种族划分的员工人数。

data <- matrix(c(6073,1033,1711,3920,3431,2178,357,757,301,332,4204,
              364,1006,337,553,7352,690,1356,1910,2066,4695,776,
              1267,575,454,3761,352,529,130,658,5523,468,652,146,
              312,5027,657,356,107,804,4650,311,674,78,599,4581,
              192,581,114,335,1176,65,121,67,195,3841,274,289,71,
              425,6489,1912,1784,1041,1434,1487,148,121,62,72,
              4130,170,365,353,479,5181,2260,1023,219,502,1286,
              1288,890,423,285,2536,289,359,61,424,6237,1504,
              1117,179,911),ncol=5,byrow=TRUE)
colnames(data) <- c("White","Black","Hispanic","Asian","Unknown")
rownames(data) <- c("A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S")
data <- as.table(data)

我正在尝试使用 R 中的 pairwise.prop.table 测试来测试公司种族的差异，看看哪些差异显着。

当我运行： pairwise.prop.test(数据[c("White","Black","Hispanic","Asian","Unknown")])

我得到 "Error in pairwise.prop.test(smoke[, c("WHITE_COUNT", "BLACK_COUNT", "HISP_COUNT", 'x' 必须有 2 列"

还有其他功能可以使用吗？我想比较每对公司的所有 5 场比赛。

如有任何帮助，我将不胜感激。谢谢！

Answer 1

正如成对文档所说，您的数据必须是

Vector of counts of successes or a matrix with 2 columns giving the counts of successes and failures, respectively

如果按照错误中的说明将列数减少到两列，就会得到一个结果。

pairwise.prop.test(data[,c("White","Black")])

将导致：

    Pairwise comparisons using Pairwise comparison of proportions 

data:  data[, c("White", "Black")] 

  A       B       C       D       E       F       G       H       I       J       K       L       M       N      
B 1.00000 -       -       -       -       -       -       -       -       -       -       -       -       -      
C < 2e-16 3.2e-14 -       -       -       -       -       -       -       -       -       -       -       -      
D < 2e-16 6.1e-14 1.00000 -       -       -       -       -       -       -       -       -       -       -      
E 1.00000 1.00000 < 2e-16 < 2e-16 -       -       -       -       -       -       -       -       -       -      
F < 2e-16 1.2e-10 1.00000 1.00000 2.8e-15 -       -       -       -       -       -       -       -       -      
G < 2e-16 < 2e-16 1.00000 1.00000 < 2e-16 1.00000 -       -       -       -       -       -       -       -      
H 4.2e-05 0.04460 1.2e-07 5.2e-07 0.00159 7.6e-05 5.6e-10 -       -       -       -       -       -       -      
I < 2e-16 < 2e-16 0.04410 8.2e-05 < 2e-16 0.00152 0.05631 < 2e-16 -       -       -       -       -       -      
J < 2e-16 < 2e-16 8.0e-14 < 2e-16 < 2e-16 < 2e-16 4.1e-14 < 2e-16 3.4e-05 -       -       -       -       -      
K < 2e-16 6.1e-14 0.04410 0.00308 1.0e-15 0.00616 0.05631 3.6e-09 1.00000 1.00000 -       -       -       -      
L < 2e-16 < 2e-16 0.50026 0.00834 < 2e-16 0.04410 0.70329 3.3e-14 1.00000 2.0e-06 1.00000 -       -       -      
M < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16 -       -      
N 3.7e-07 6.8e-05 1.00000 1.00000 4.2e-06 1.00000 1.00000 0.12875 0.00597 5.4e-13 0.00571 0.05631 < 2e-16 -      
O < 2e-16 < 2e-16 2.0e-13 < 2e-16 < 2e-16 2.5e-16 1.2e-13 < 2e-16 3.4e-05 1.00000 1.00000 2.1e-06 < 2e-16 7.2e-13
P < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16
Q < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16
R 8.3e-07 0.00079 0.03436 0.23508 2.0e-05 0.48752 0.00659 1.00000 2.4e-08 < 2e-16 1.4e-05 5.8e-06 < 2e-16 1.00000
S 2.1e-13 9.6e-08 < 2e-16 < 2e-16 3.2e-13 < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16 < 2e-16 1.3e-05 < 2e-16
  O       P       Q       R      
B -       -       -       -      
C -       -       -       -      
D -       -       -       -      
E -       -       -       -      
F -       -       -       -      
G -       -       -       -      
H -       -       -       -      
I -       -       -       -      
J -       -       -       -      
K -       -       -       -      
L -       -       -       -      
M -       -       -       -      
N -       -       -       -      
O -       -       -       -      
P < 2e-16 -       -       -      
Q < 2e-16 < 2e-16 -       -      
R < 2e-16 < 2e-16 < 2e-16 -      
S < 2e-16 < 2e-16 < 2e-16 < 2e-16

P value adjustment method: holm

Answer 2

我希望这可以是在黑暗中拍摄的。通过这种方式，您应该能够针对每个种族对公司之间的成对比较进行比较。事实上，您需要在多项分布之间执行多重比较。脚步： - 数据从宽格式转换为长格式； - Poisson GLM 以频率作为结果，公司和种族作为协变量； - emmeans 包用于成对比较最终输出是每场比赛的公司之间的对数赔率差异。

data <- matrix(c(6073,1033,1711,3920,3431,2178,357,757,301,332,4204,
                 364,1006,337,553,7352,690,1356,1910,2066,4695,776,
                 1267,575,454,3761,352,529,130,658,5523,468,652,146,
                 312,5027,657,356,107,804,4650,311,674,78,599,4581,
                 192,581,114,335,1176,65,121,67,195,3841,274,289,71,
                 425,6489,1912,1784,1041,1434,1487,148,121,62,72,
                 4130,170,365,353,479,5181,2260,1023,219,502,1286,
                 1288,890,423,285,2536,289,359,61,424,6237,1504,
                 1117,179,911),ncol=5,byrow=TRUE)
colnames(data) <- c("White","Black","Hispanic","Asian","Unknown")
rownames(data) <- c("A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S")
data
typeof(data)
data <- as.data.frame(data)

library(tidyverse)
data2 <- data %>%
  rownames_to_column(var="Firm") %>%
  gather(key = Race, value = "n", White:Unknown, factor_key=F)
data2
fit <- glm(n ~ Firm+Race, data = data2, family = poisson)
fit
library(emmeans)
pairs(emmeans(fit, ~ Firm|Race))

pairwise.prop.test 在具有多个类别的 R 中

pairwise.prop.test in R with Multiple Categories

comparison

r

pairwise