我正在尝试将multcompView信函与摘要数据连接到一个矩阵中。我使用一个for loop来运行单独的摘要矩阵单元格,并将它们与它们各自的字母连接起来。我快到了,但我的矩阵输出原始数据和粘贴数据(见下文)。
功能:
for (i in 1:nrow(X1))
tableRow = matrix(c(tableRow,paste(tableRow[i],letters$Letters[i],sep = "")),nrow = 1)其中:X1是我的汇总表,tableRow是X1的第一行,letters包含我想连接的字母。
返回:
[1,] "5.53 ± 0.77" "6.72 ± 1.18" "5.12 ± 0.44"
"5.24 ± 0.41" "5.53 ± 0.77a" "6.72 ± 1.18a" "5.12 ± 0.44a" "5.24 ± 0.41a"期望产出:
[1,] "5.53 ± 0.77a" "6.72 ± 1.18a" "5.12 ± 0.44a" "5.24 ± 0.41a"发布于 2016-06-24 01:51:03
这将完成所有的工作:
## example matrix
set.seed(0); X <- round(matrix(runif(12), nrow = 4, ncol = 3), 2)
# [,1] [,2] [,3]
# [1,] 0.90 0.91 0.66
# [2,] 0.27 0.20 0.63
# [3,] 0.37 0.90 0.06
# [4,] 0.57 0.94 0.21
matrix(paste0(X, letters[1:4]), nrow = nrow(X))
# [,1] [,2] [,3]
# [1,] "0.9a" "0.91a" "0.66a"
# [2,] "0.27b" "0.2b" "0.63b"
# [3,] "0.37c" "0.9c" "0.06c"
# [4,] "0.57d" "0.94d" "0.21d"对于您的数据,您可以:
matrix(paste0(X1, letters$Letters), nrow = nrow(X1))备注1
我的例子有一些缺陷。您已经将X1作为字符矩阵,而我的示例X是数字的。执行paste0()时,数值0.90变为"0.9" (因为as.character(0.90)给出了"0.9")。对于你的数据,不会有这样的行为。
备注2
哦,我真的找到了避免这种行为的方法。
X <- format(X)
# [,1] [,2] [,3]
# [1,] "0.90" "0.91" "0.66"
# [2,] "0.27" "0.20" "0.63"
# [3,] "0.37" "0.90" "0.06"
# [4,] "0.57" "0.94" "0.21"那么做paste0()就可以了:
# [,1] [,2] [,3]
# [1,] "0.90a" "0.91a" "0.66a"
# [2,] "0.27b" "0.20b" "0.63b"
# [3,] "0.37c" "0.90c" "0.06c"
# [4,] "0.57d" "0.94d" "0.21d"https://stackoverflow.com/questions/38004288
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