Question

作为预测包的checkresiduals（）函数和rbind（）函数的结果，我得到了此矩阵（ETS_RESIDUALS）：

#Result of checkresiduals() function

     [,1]                                         
[1,] "Q* = 161.83, df = 18.8, p-value < 2.2e-16"  
[2,] "Q* = 125.46, df = 18.8, p-value < 2.2e-16"  
[3,] "Q* = 263.65, df = 18.8, p-value < 2.2e-16"  
[4,] "Q* = 81.503, df = 18.8, p-value = 8.763e-10"
[5,] "Q* = 36.616, df = 18.8, p-value = 0.008178" 

str(ETS_RESIDUALS)
#chr [1:5, 1] "Q* = 161.83, df = 18.8, p-value < 2.2e-16" "Q* = 125.46, df = 18.8, p-value < 2.2e-16" "Q* = 263.65, df = 18.8, p-value < 2.2e-16" ...

class(ETS_RESIDUALS)
#[1] "matrix"

现在，我的意图是使用grep（）或其他函数将这行文本拆分为一个data.frame（具有四列TEST，Q *，df，p值），如下例所示：

TEST      Q*        df        p-value
--------------------------------------------
TEST_1  161.83     18.8        2.2e-16  
TEST_2  125.46     18.8        2.2e-16  
TEST_3  263.65     18.8        2.2e-16  
TEST_4  81.503     18.8        8.763e-10
TEST_5  36.616     18.8        0.008178

我尝试使用这行代码，但是效果不好。

ETS_RESIDUALS %>%   
  stringr::str_replace_all("(\\S+) =", "`\\1` =") %>%   
  paste0("data.frame(", ., ", check.names = FALSE)")

有人可以用这个代码帮助我吗？

Answer 1

library(dplyr)
library(tidyr)
library(stringr)
#separate based on ,
separate(data.frame(mat), mat ,into = c('Q*','df','p-value'),sep = ',') %>% 
mutate_all(~str_extract(.,'(?<=\\=|\\<\\s).*')) %>% 
#Use positive look-behind to extract everything after = or < followed by a space
mutate(TEST=paste0('TEST_',1:n())) %>% select(TEST,everything())

    TEST      Q*    df    p-value
1 TEST_1  161.83  18.8    2.2e-16
2 TEST_2  125.46  18.8    2.2e-16
3 TEST_3  263.65  18.8    2.2e-16
4 TEST_4  81.503  18.8  8.763e-10
5 TEST_5  36.616  18.8   0.008178

数据

mat <- structure(c("Q* = 161.83, df = 18.8, p-value < 2.2e-16", "Q* = 125.46, df = 18.8, p-value < 2.2e-16", "Q* = 263.65, df = 18.8, 
       p-value < 2.2e-16", "Q* = 81.503, df = 18.8, p-value = 8.763e-10", "Q* = 36.616, df = 18.8, p-value = 0.008178"), 
      .Dim = c(5L, 1L))

Answer 2

您可以将strsplit用于array，然后求解为data.frame对象。

A <- array(apply(M, 1, function(x) unlist(strsplit(strsplit(x, ", ")[[1]], "\\s[<=]\\s"))), 
  c(2, 3, nrow(M)))
d <- setNames(as.data.frame(t(apply(A, 3, function(x) as.numeric(x[2, ])))), A[1,,1])

产量

d
#        Q*   df   p-value
# 1 161.830 18.8 2.200e-16
# 2 125.460 18.8 2.200e-16
# 3 263.650 18.8 2.200e-16
# 4  81.503 18.8 8.763e-10
# 5  36.616 18.8 8.178e-03

数据

M <- structure(c("Q* = 161.83, df = 18.8, p-value < 2.2e-16", "Q* = 125.46, df = 18.8, 
                 p-value < 2.2e-16", 
"Q* = 263.65, df = 18.8, p-value < 2.2e-16", "Q* = 81.503, df = 18.8, p-value = 8.763e-10", 
"Q* = 36.616, df = 18.8, p-value = 0.008178"), .Dim = c(5L, 1L
))

Answer 3

一种方法是拆分和使用逻辑子集。

v1 <- unlist(strsplit(m1, ' = | < | > |, '))
setNames(as.data.frame(matrix(v1[c(FALSE, TRUE)], nrow = nrow(m1), byrow = TRUE)), 
                                                                   unique(v1[c(TRUE, FALSE)]))
#      Q*   df p-value
#1 161.83 18.8 2.2e-16
#2 161.83 18.8 2.2e-16
#3 161.83 18.8 2.2e-16
#4 161.83 18.8 2.2e-16
#5 161.83 18.8 2.2e-16

要创建额外的列，只需

df$TEST <- paste0('TEST', seq(nrow(df)))

注意：所有值都是相同的，因为样本矩阵的所有字符串都相同，

[,1]                                       
[1,] "Q* = 161.83, df = 18.8, p-value < 2.2e-16"
[2,] "Q* = 161.83, df = 18.8, p-value < 2.2e-16"
[3,] "Q* = 161.83, df = 18.8, p-value < 2.2e-16"
[4,] "Q* = 161.83, df = 18.8, p-value < 2.2e-16"
[5,] "Q* = 161.83, df = 18.8, p-value < 2.2e-16"

提取文本并放入表格

3 个答案: