合并数据框和覆盖值

Question

如何合并2个相似的数据框,但有一个更重要的数据框？

例如:

数据帧1

Date      Col1    Col2
jan         2      1
feb         4      2
march       6      3
april       8      NA

数据帧2

Date      Col2    Col3
jan         9      10
feb         8      20
march       7      30
april       6      40

将这些按日期合并,数据框1优先,但数据框2填充空白

DataframeMerge

Date      Col1    Col2    Col3
jan         2       1      10
feb         4       2      20
march       6       3      30
april       8       6      40

编辑 - 解决方案

commonNames <- names(df1)[which(colnames(df1) %in% colnames(df2))]
commonNames <- commonNames[commonNames != "key"]
dfmerge<- merge(df1,df2,by="key",all=T)
for(i in commonNames){
  left <- paste(i, ".x", sep="")
  right <- paste(i, ".y", sep="")
  dfmerge[is.na(dfmerge[left]),left] <- dfmerge[is.na(dfmerge[left]),right]
  dfmerge[right]<- NULL
  colnames(dfmerge)[colnames(dfmerge) == left] <- i
}

Answer 1

merdat <- merge(dfrm1,dfrm2, by="Date")  # seems self-documenting

#  explanation for next line in text below.
merdat$Col2.y[ is.na(merdat$Col2.y) ] <- merdat$Col2.x[ is.na(merdat$Col2.y) ]

然后只需将'merdat $ Col2.y'重命名为'merdat $ Col2'并删除'merdat $ Col2.x'.

在回复请求更多注释时:仅更新向量的各个部分的一种方法是构造用于索引的逻辑向量,并使用"["将其应用于赋值的两侧.另一种方法是设计一个逻辑向量,该逻辑向量仅在赋值的LHS上,但随后使用rep()具有相同长度的向量sum(logical.vector).目标是两个实例都具有与被替换项目相同的长度(和顺序).

Answer 2

使用data.table的on=参数v1.9.6更新(允许adhoc连接:

setDT(df1)[df2, `:=`(Col2 = ifelse(is.na(Col2), i.Col2, Col2), 
                     Col3 = i.Col3), on="Date"][]

这是一个data.table解决方案.确保您df1和df2的Date列因素与预期水平(订货)

require(data.table)
dt1 <- data.table(df1, key="Date")
dt2 <- data.table(df2, key="Date")
# Col2 refers to the Col2 of dt1 and i.col2 refers to that of dt2
dt1[dt2, `:=`(Col3 = Col3, Col1 = Col1, 
        Col2 = ifelse(is.na(Col2), i.Col2, Col2))]

# the result is stored in dt1
> dt1
#     Date Col1 Col2 Col3
# 1:   jan    2    1   10
# 2:   feb    4    2   20
# 3: march    6    3   30
# 4: april    8    6   40

Answer 3

这是一个dplyr解决方案.感谢@docendo discimus

df1 <- data.frame(y = c("A", "B", "C", "D"), x1 = c(1,2,NA, 4)) 

  y x1
1 A  1
2 B  2
3 C NA
4 D  4

df2 <- data.frame(y = c("A", "B", "C"), x1 = c(5, 6, 7))

  y x1
1 A  5
2 B  6
3 C  7

dplyr

left_join(df1, df2, by="y") %>% 
transmute(y, x1 = ifelse(is.na(x1.y), x1.x, x1.y))

  y x1
1 A  5
2 B  6
3 C  7

Answer 4

考虑这个例子:

> d1 <- data.frame(x=1:4, a=2:5, b=c(3,4,5,NA))
> d1
  x a  b
1 1 2  3
2 2 3  4
3 3 4  5
4 4 5 NA
> d2 <- data.frame(x=1:4, b=c(6,7,8,9), c=11:14)
> d2
  x b  c
1 1 6 11
2 2 7 12
3 3 8 13
4 4 9 14

现在使用merge和within,用ifelse:

> within(merge(d1, d2, by="x"), {b <- ifelse(is.na(b.x),b.y,b.x); b.x <- NULL; b.y <- NULL})
  x a  c b
1 1 2 11 3
2 2 3 12 4
3 3 4 13 5
4 4 5 14 9