Question

我有以下数据

df<-structure(list(X__1 = c("Q6NVC0 [197-213]", "A0A0A8UAS1 [48-64]", 
"A0A0F0XK65 [8-17]", "A0A0G2JMX7 [481-491]; P20406-6 [106-116]; Q6UIN5 [7-17]; P20406-3 [70-80]; P20406-8 [164-174]", 
"A0A0G2JMX7 [481-497]; P20406-6 [106-122]; Q6UIN5 [7-23]; P20406-3 [70-86]; P20406-8 [164-180]", 
"P20406-6 [106-132]; Q6UIN5 [7-33]; P20406-3 [70-96]; P20406-8 [164-190]", 
"H6VRG2 [603-616]", "P13645 [41-59]", "P35527 [488-513]", "P35908 [525-544]; H2R1Z0 [512-531]", 
"H6VRG2 [550-588]", "A0A024RBS2 [150-162]", "A0A023L3M5 [83-110]", 
"P10809 [143-156]", "V9HW22 [470-493]", "P20406-8 [379-392]", 
"P19338 [524-537]", "A0A024L4P7 [46-57]", "G7YYH5 [45-55]", "Q86YZ3 [973-991]; [1443-1461]; [1913-1931]; [2383-2401]", 
"P35527 [375-390]", "P35527 [375-391]", "D2TTS3 [41-50]")), .Names = "X__1", class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -23L))

我正在尝试制作这样的输出首先，我尝试获取两个ID P20406-8和A0A024RBS2旁边的值。如果它们不存在，我们将获取任何其他ID的前两个值。如果它们一起存在，我们取第一个值旁边的值，如果其中一个ID存在，我们就取其值。

所以最后我想要两列值，如下面的

Answer 1

这段代码对我来说很合适：

df_new <- df2$X__1
primary_ID <- "P20406-8"
secondary_ID <- "A0A024RBS2"

ID_Primary <- grepl(pattern = primary_ID, df_new)
ID_Secondary <- grepl(pattern = secondary_ID, df_new)

df_new[ID_Primary] <- substr(df_new[ID_Primary],
                              start = regexpr(primary_ID, df_new[ID_Primary]), 
                              stop = 2000)

df_new[ID_Secondary & !ID_Primary] <- substr(df_new[ID_Secondary],
                                               start = regexpr(secondary_ID, df_new[ID_Secondary]), 
                                               stop = 2000)

start <- regexpr(pattern = "\\[", text = df_new)  # find opening square bracket
end <- regexpr(pattern = "\\]", text = df_new)  # find closing square bracket
temp <- substr(df_new, start+1, end-1)  # only take values between opening and closing
temp <- strsplit(temp, "-")  # split values in two

result <- t(data.frame(temp))
rownames(result) <- NULL

我希望这有帮助！

Answer 2

一种方法：

library(splitstackshape)
# separate multi observation lines into multiple columns
df2 <- cSplit(df, splitCols='X__1', sep=';')
# put in a long format for easier processing
df3 <- na.omit(data.frame(melt(as.data.frame(df2),measure=names(df2))))

# extractions
# account id
df3$id <- with(df3,gsub(x=value,pattern="^(.*) [[:punct:]](.*)-(.*)[[:punct:]]$",replacement='\\1'))
# start range
df3$begin <- with(df3,gsub(x=value,pattern="^(.*) [[:punct:]](.*)-(.*)[[:punct:]]$",replacement='\\2'))
# end range
df3$end <- with(df3,gsub(x=value,pattern="^(.*) [[:punct:]](.*)-(.*)[[:punct:]]$",replacement='\\3'))
# accounts with multiple:
df3$mult_id <- as.numeric(with(df3,gsub(x=variable,pattern="^X__1_(.*)$",replacement='\\1')))


# apparently you only want those with one obs:
df.final <- df3[which(df4$mult_id==1),]

如何从多个字符串中取值

2 个答案: