我有以下三个数据帧:
df1 <- data.frame(name=c("John", "Anne", "Christine", "Andy"),
age=c(31, 26, 54, 48),
height=c(180, 175, 160, 168),
group=c("Student",3,5,"Employer"), stringsAsFactors=FALSE)
df2 <- data.frame(name=c("Anne", "Christine"),
age=c(26, 54),
height=c(175, 160),
group=c(3,5),
group2=c("Teacher",6), stringsAsFactors=FALSE)
df2 <- data.frame(name=c("Christine"),
age=c(54),
height=c(160),
group=c(5),
group2=c(6),
group3=c("Scientist"), stringsAsFactors=FALSE)
我想将它们组合起来,以便得到以下结果:
df.all <- data.frame(name=c("John", "Anne", "Christine", "Andy"),
age=c(31, 26, 54, 48),
height=c(180, 175, 160, 168),
group=c("Student", "Teacher", "Scientist", "Employer"))
目前我正是这样做的:
df.all <- merge(merge(df1[,c(1,4)], df2[,c(1,5)], all=TRUE, by="name"),
df3[,c(1,6)], all=TRUE, by="name")
row.ind <- which(df.all$group %in% c(6,5))
df.all[row.ind, c("group")] <- df.all[row.ind, c("group2")]
row.ind2 <- which(df.all$group2 %in% c(6))
df.all[row.ind2, c("group")] <- df.all[row.ind2, c("group3")]
这不是一般性的,而且非常混乱。也许有一种方法可以使用merge_all
或merge_recurse
进行合并步骤(特别是因为可能有两个以上的数据框要合并),但我还没弄清楚如何。这两个不能产生正确的结果:
df.all <- merge_all(list(df1, df2, df3))
df.all <- merge_recurse(list(df1, df2, df3), by=c("name"))
是否有更通用,更优雅的方法来解决这个问题?
答案 0 :(得分:5)
如果我明白你最终会追求什么,这是另一种可行的方法。 (目前尚不清楚“组”列中的数值是什么,所以我不确定这正是您正在寻找的。)
使用Reduce()
合并您的多个data.frame
。
temp <- Reduce(function(x, y) merge(x, y, all=TRUE), list(df1, df2, df3))
names(temp)[4] <- "group1" # Rename "group" to "group1" for reshaping
temp
# name age height group1 group2 group3
# 1 Andy 48 168 Employer <NA> <NA>
# 2 Anne 26 175 3 Teacher <NA>
# 3 Christine 54 160 5 6 Scientist
# 4 John 31 180 Student <NA> <NA>
使用reshape()
从长到长重塑您的数据。
df.all <- reshape(temp, direction = "long", idvar="name", varying=4:6, sep="")
df.all
# name age height time group
# Andy.1 Andy 48 168 1 Employer
# Anne.1 Anne 26 175 1 3
# Christine.1 Christine 54 160 1 5
# John.1 John 31 180 1 Student
# Andy.2 Andy 48 168 2 <NA>
# Anne.2 Anne 26 175 2 Teacher
# Christine.2 Christine 54 160 2 6
# John.2 John 31 180 2 <NA>
# Andy.3 Andy 48 168 3 <NA>
# Anne.3 Anne 26 175 3 <NA>
# Christine.3 Christine 54 160 3 Scientist
# John.3 John 31 180 3 <NA>
利用as.numeric()
将字符强制转换为NA
这一事实,并使用na.omit()
删除所有NA
个值的行。
na.omit(df.all[is.na(as.numeric(df.all$group)), ])
# name age height time group
# Andy.1 Andy 48 168 1 Employer
# John.1 John 31 180 1 Student
# Anne.2 Anne 26 175 2 Teacher
# Christine.3 Christine 54 160 3 Scientist
同样,这可能会过度概括您的问题 - 例如,其他列中可能存在NA值 - 但它可能有助于指导您解决问题。
答案 1 :(得分:4)
第一步是将merge_recurse
与all.x = TRUE
:
library(reshape)
merge.all <- merge_recurse(list(df1, df2, df3), all.x = TRUE)
# name age height group group2 group3
# 1 Anne 26 175 3 Teacher <NA>
# 2 Christine 54 160 5 6 Scientist
# 3 John 31 180 Student <NA> <NA>
# 4 Andy 48 168 Employer <NA> <NA>
然后,您可以使用apply
从所有“组”列中获取最后一个非NA
组:
group.cols <- grep("group", colnames(merge.all))
merge.all <- data.frame(merge.all[-group.cols],
group = apply(merge.all[group.cols], 1,
function(x)tail(na.omit(x), 1)))
# name age height group
# 1 Anne 26 175 Teacher
# 2 Christine 54 160 Scientist
# 3 John 31 180 Student
# 4 Andy 48 168 Employer