从向量R中提取带有十进制的数字

Question

我有以下向量：

c("c(`Kruskal-Wallis chi-squared` = 201.760850624131)", "c(df = 17)", 
"1.26686891197831e-33", "Kruskal-Wallis rank sum test", "delta_Z by criteria"
)

我想要这个输出：

c("201.760850624131", "17", "1.26686891197831e-33")

谢谢你的帮助

Answer 1

我们可以将str_extract与匹配一个或多个数字([0-9]+)的正则表达式([0-9]+)与一个或多个数字(然后是一个或多个数字)匹配，e与-或+以及任何数字相匹配。

library(stringr)
as.numeric(na.omit(str_extract(v1, "[0-9]+(\\.[0-9]+e[-+]\\d+)?")))

-output

[1] 2.010000e+02 1.700000e+01 1.266869e-33

数据

v1 <- c("c(`Kruskal-Wallis chi-squared` = 201.760850624131)", "c(df = 17)", 
"1.26686891197831e-33", "Kruskal-Wallis rank sum test", "delta_Z by criteria"
)

Answer 2

我们可以尝试在这里使用str_extract作为regex选项：

x <- c("c(`Kruskal-Wallis chi-squared` = 201.760850624131)",
       "c(df = 17)", 
       "1.26686891197831e-33",
       "Kruskal-Wallis rank sum test",
       "delta_Z by criteria"
)
output <- as.numeric(na.omit(str_extract(x, "\\d+(?:\\.\\d+)?(?:e[+-]\\d+)?")))
output

[1] 2.017609e+02 1.700000e+01 1.266869e-33

对regex的解释：

• \\d+匹配整数component
• (?:\\.\\d+)?可选的十进制component
• (?:e[+-]\\d+)?可选指数(正或负)

Answer 3

因为它似乎是R码：

x <- c("c(`Kruskal-Wallis chi-squared` = 201.760850624131)", "c(df = 17)", 
"1.26686891197831e-33", "Kruskal-Wallis rank sum test", "delta_Z by criteria")

as.character(sapply(sapply(x[1:3], str2lang, USE.NAMES = F), eval))

#> [1] "201.760850624131"     "17"                   "1.26686891197831e-33"