R中基于行值和类别的条件计算答案

【问题标题】：Conditional calculation in R based on Row values and categoriesR中基于行值和类别的条件计算
【发布时间】：2017-10-11 14:51:28
【问题描述】：

我有这个数据框：

df<-data.frame(a=c("a1","a2","a3","a4","b1","b2","b3","b4","a1","a2","a3","a4","b1","b2","b3","b4"), b=c("x1","x2","x3","total","x1","x2","x3","total", "x1","x2","x3","total","x1","x2","x3","total"), reg=c("A","A","A","A","A","A","A","A","B", "B","B","B","B","B","B","B"), c=c(1:16))

看起来像：

    a     b reg  c
1  a1    x1   A  1
2  a2    x2   A  2
3  a3    x3   A  3
4  a4 total   A  4
5  b1    x1   A  5
6  b2    x2   A  6
7  b3    x3   A  7
8  b4 total   A  8
9  a1    x1   B  9
10 a2    x2   B 10
11 a3    x3   B 11
12 a4 total   B 12
13 b1    x1   B 13
14 b2    x2   B 14
15 b3    x3   B 15
16 b4 total   B 16

列“a”、“b”和“reg”是分类变量。我想要做的是创建一个新列，它将 x(i) 相除，其中 i=1,2,3 与 reg' and ina' 列中每个类别的“总计”(x(i)/total)。

有人可以帮我吗？

【问题讨论】：

标签： r data-manipulation categorical-data calculation

【解决方案1】：

假设您的 df 像您的示例一样订购。

library(zoo)
df$NEW=df$c
df$NEW[df$b!='total']=NA
df$NEW=na.locf(df$NEW,fromLast=T,na.rm=F)
df$NEW=df$c/df$NEW

df
    a     b reg  c       NEW
1  a1    x1   A  1 0.2500000
2  a2    x2   A  2 0.5000000
3  a3    x2   A  3 0.7500000
4  a4 total   A  4 1.0000000
5  b1    x1   A  5 0.6250000
6  b2    x2   A  6 0.7500000
7  b3    x2   A  7 0.8750000
8  b4 total   A  8 1.0000000
9  a1    x1   B  9 0.7500000
10 a2    x2   B 10 0.8333333
11 a3    x2   B 11 0.9166667
12 a4 total   B 12 1.0000000
13 b1    x1   B 13 0.8125000
14 b2    x2   B 14 0.8750000
15 b3    x2   B 15 0.9375000
16 b4 total   B 16 1.0000000

根据Op的解释，以下是他/她的真实数据。（来自OP）

data1$shares<-NA 
id<-which(data1$Occupation=='Total') 
data1$shares[id]<-data1$2014[id]
data1$shares=na.locf(data1$shares,fromLast=T,na.rm=F) 
data1$shares=data1$2014/data1$shares

【讨论】：

由于某种原因没有工作。是因为在某些情况下total 是NAN，而“c”列可能还包含“NAN”值？
@user17880 如果 total 是 NA ，你应该如何处理
理想情况下，在本例中，它应该返回 'NA'。
仍然没有工作。但是，我发现对您的代码的这种修改符合我的目的： 'data1$shares2014 [id] data1$shares=na.locf(data1$shares,fromLast=T,na.rm=F) data1$shares=data1$2014/data1$shares'
不，最好自己修改答案。我认为你应该得到大部分的功劳:)

【解决方案2】：

仅使用 R 基础：

df<-data.frame(a=c("a1","a2","a3","a4","b1","b2","b3","b4","a1","a2","a3","a4","b1","b2","b3","b4"), b=c("x1","x2","x3","total","x1","x2","x3","total", "x1","x2","x3","total","x1","x2","x3","total"), reg=c("A","A","A","A","A","A","A","A","B", "B","B","B","B","B","B","B"), c=c(1:16))

totals <- data.frame(To=df[df$b=='total',4])
totals$from <- c(1, totals$To[1:nrow(totals)-1]+1)
df$NEW = df$c/totals[findInterval(x=df$c, vec=c(rbind(totals$from, totals$to))), 1]
df

输出：

    a     b reg  c       NEW
1  a1    x1   A  1 0.2500000
2  a2    x2   A  2 0.5000000
3  a3    x3   A  3 0.7500000
4  a4 total   A  4 1.0000000
5  b1    x1   A  5 0.6250000
6  b2    x2   A  6 0.7500000
7  b3    x3   A  7 0.8750000
8  b4 total   A  8 1.0000000
9  a1    x1   B  9 0.7500000
10 a2    x2   B 10 0.8333333
11 a3    x3   B 11 0.9166667
12 a4 total   B 12 1.0000000
13 b1    x1   B 13 0.8125000
14 b2    x2   B 14 0.8750000
15 b3    x3   B 15 0.9375000
16 b4 total   B 16 1.0000000

【讨论】：