用查找表中的匹配项替换dataframe中的每一列
我有一个用查找表中的匹配项替换dataframe中的每一列,r,dataframe,lookup,R,Dataframe,Lookup,我有一个data.frame称为table_1，结构如下： p_id rd1 rd2 rd3 <fctr><fctr><fctr><fctr> 1 1 5 4 6 2 2 3 1 1 3 3 6 6 5 4 4 1 5 2 5 5 4 1
data.frametable_1 p_id rd1 rd2 rd3
<fctr><fctr><fctr><fctr>
1 1 5 4 6
2 2 3 1 1
3 3 6 6 5
4 4 1 5 2
5 5 4 1 4
表1p\u scrrd1rd2rd3 p_id rd1 rd2 rd3
<fctr><fctr><fctr><fctr>
1 1 55 44 66
2 2 33 11 11
3 3 66 66 55
4 4 11 55 22
5 5 44 11 44
p_id rd1 rd2 rd3
1 1 55 44 66
2 2 33 11 11
3 3 66 66 55
4 4 11 55 22
5 5 44 11 44
我怀疑这将使用
mappylappymatchmutate# t is your df; ref is your lookup table
t <- data.frame(p_id=factor(c(10,20,30,40,50)),
rd1=factor(c(5,3,6,1,4)*10),
rd2=factor(c(4,1,6,5,1)*10),
rd3=factor(c(6,1,5,2,4)*10))
ref <- data.frame(p_id=factor(c(10,20,30,40,50,60)),
p_scr=factor(c(11,22,33,44,55,66)))
t
# p_id rd1 rd2 rd3
# 1 10 50 40 60
# 2 20 30 10 10
# 3 30 60 60 50
# 4 40 10 50 20
# 5 50 40 10 40
ref
# p_id p_scr
# 1 10 11
# 2 20 22
# 3 30 33
# 4 40 44
# 5 50 55
# 6 60 66
# assuming p_id is unique, set rownames of ref lookup table to p_id to allow for indexing by p_id
rownames(ref) <- ref$p_id
rownames(ref) # character values, not numeric
# [1] "10" "20" "30" "40" "50" "60"
# ref lookup table now looks like this
ref
# p_id p_scr
# 10 10 11
# 20 20 22
# 30 30 33
# 40 40 44
# 50 50 55
# 60 60 66
# single case, ref rownames are character vectors, we want to index with corresponding character vector from t
as.character(t$rd1)
# [1] "50" "30" "60" "10" "40"
ref[as.character(t$rd1),]$p_scr # use character values of rd1 to index, matching the character values of rownames
# [1] 55 33 66 11 44
# Levels: 11 22 33 44 55 66
# apply to each rd column, returns the character values of p_scr factor
apply(t[,2:ncol(t)], 2, function(x) ref[as.character(x),]$p_scr)
# converts to numeric the character values of p_scr factor
apply(t[,2:ncol(t)], 2, function(x) as.numeric(as.character(ref[as.character(x),]$p_scr)))
# NOTE: the previous answer I gave does not work, why?
ref[t$rd1,]$p_scr # gives incorrect order
# [1] 44 22 55 11 33
# Levels: 11 22 33 44 55 66
# NOTE structure of t
str(t)
# 'data.frame': 5 obs. of 4 variables:
# $ p_id: Factor w/ 5 levels "10","20","30",..: 1 2 3 4 5
# $ rd1 : Factor w/ 5 levels "10","30","40",..: 4 2 5 1 3
# $ rd2 : Factor w/ 4 levels "10","40","50",..: 2 1 4 3 1
# $ rd3 : Factor w/ 5 levels "10","20","40",..: 5 1 4 2 3
# Do you see the character vs integer values of the factor t$rd1
t$rd1
# [1] 50 30 60 10 40
# Levels: 10 30 40 50 60
# The levels of t$rd1: "10", "30", "40", "50", "60", which correspond to 4 2 5 1 3 position
# In the case of ref[t$rd1] you are using the integer values of t$rd1 and indexing ref by position: ref[c(4,2,5,1,3)] so your output is c(44, 22, 55, 11, 33)
# In the case of ref[as.character(t$rd1) you are using the character values of t$rd1 and indexing ref by rownames: ref[c("50", "30", "60", "10", "40")] so your output is c(55, 33, 66 11, 44)
#t是您的df；ref是您的查找表
这个问题还不完全清楚，但是如果你想了解apply函数族的入门知识，请参考R base软件包，了解所有函数及其示例的详细用法。令人惊叹的太棒了。这起作用令人惊讶。我想我还是不明白为什么rownames（ref）
和允许索引在这里工作。我将不得不继续关注单一用例（感谢添加该用例）。谢谢大家!@jmb277刚刚编辑添加了警告，如果您的数据是因子，请注意因子的级别。我做了一些抽样检查，它似乎有效。我认为数据仍然是错误的。我想我不清楚这是否会成为一个问题。apply
的结果要么是一个数字，要么是一个NA，这对我来说没问题。嗯，是的，我想我刚才遇到了这个问题。当我尝试将查找到的数据转换为数字时，它会被更改为我无法识别的数字。数字转换可以在这个部分中完成吗<代码>应用（t[，2:ncol（t）]，2，函数（x）ref[x，]$p_scr）# t is your df; ref is your lookup table
t <- data.frame(p_id=factor(c(10,20,30,40,50)),
rd1=factor(c(5,3,6,1,4)*10),
rd2=factor(c(4,1,6,5,1)*10),
rd3=factor(c(6,1,5,2,4)*10))
ref <- data.frame(p_id=factor(c(10,20,30,40,50,60)),
p_scr=factor(c(11,22,33,44,55,66)))
t
# p_id rd1 rd2 rd3
# 1 10 50 40 60
# 2 20 30 10 10
# 3 30 60 60 50
# 4 40 10 50 20
# 5 50 40 10 40
ref
# p_id p_scr
# 1 10 11
# 2 20 22
# 3 30 33
# 4 40 44
# 5 50 55
# 6 60 66
# assuming p_id is unique, set rownames of ref lookup table to p_id to allow for indexing by p_id
rownames(ref) <- ref$p_id
rownames(ref) # character values, not numeric
# [1] "10" "20" "30" "40" "50" "60"
# ref lookup table now looks like this
ref
# p_id p_scr
# 10 10 11
# 20 20 22
# 30 30 33
# 40 40 44
# 50 50 55
# 60 60 66
# single case, ref rownames are character vectors, we want to index with corresponding character vector from t
as.character(t$rd1)
# [1] "50" "30" "60" "10" "40"
ref[as.character(t$rd1),]$p_scr # use character values of rd1 to index, matching the character values of rownames
# [1] 55 33 66 11 44
# Levels: 11 22 33 44 55 66
# apply to each rd column, returns the character values of p_scr factor
apply(t[,2:ncol(t)], 2, function(x) ref[as.character(x),]$p_scr)
# converts to numeric the character values of p_scr factor
apply(t[,2:ncol(t)], 2, function(x) as.numeric(as.character(ref[as.character(x),]$p_scr)))
# NOTE: the previous answer I gave does not work, why?
ref[t$rd1,]$p_scr # gives incorrect order
# [1] 44 22 55 11 33
# Levels: 11 22 33 44 55 66
# NOTE structure of t
str(t)
# 'data.frame': 5 obs. of 4 variables:
# $ p_id: Factor w/ 5 levels "10","20","30",..: 1 2 3 4 5
# $ rd1 : Factor w/ 5 levels "10","30","40",..: 4 2 5 1 3
# $ rd2 : Factor w/ 4 levels "10","40","50",..: 2 1 4 3 1
# $ rd3 : Factor w/ 5 levels "10","20","40",..: 5 1 4 2 3
# Do you see the character vs integer values of the factor t$rd1
t$rd1
# [1] 50 30 60 10 40
# Levels: 10 30 40 50 60
# The levels of t$rd1: "10", "30", "40", "50", "60", which correspond to 4 2 5 1 3 position
# In the case of ref[t$rd1] you are using the integer values of t$rd1 and indexing ref by position: ref[c(4,2,5,1,3)] so your output is c(44, 22, 55, 11, 33)
# In the case of ref[as.character(t$rd1) you are using the character values of t$rd1 and indexing ref by rownames: ref[c("50", "30", "60", "10", "40")] so your output is c(55, 33, 66 11, 44)
n <- 1:5 # numeric
n
f <- factor(n, levels=5:1) # factor
f
levels(f)
# consequence when used to index
letters[n]
[1] "a" "b" "c" "d" "e"
letters[f]
[1] "e" "d" "c" "b" "a"