用if-语句循环应用dataframe的行

I'm new to R and I'm trying to sum 2 columns of a given dataframe, if both the elements to be summed satisfy a given condition. To make things clear, what I want to do is:

我是R的新手，我试着求和一个给定的dataframe的两列，如果要求和的两个元素都满足给定的条件。为了把事情弄清楚，我想做的是:

> t.d<-as.data.frame(matrix(1:9,ncol=3))
> t.d
  V1 V2 V3
  1  4  7  
  2  5  8  
  3  6  9  

> t.d$V4<-rep(0,nrow(t.d))

> for (i in 1:nrow(t.d)){
+   if (t.d$V1[i]>1 && t.d$V3[i]<9){
+     t.d$V4[i]<-t.d$V1[i]+t.d$V3[i]}
+     }

> t.d    
  V1 V2 V3 V4
  1  4  7  0
  2  5  8 10
  3  6  9  0

I need an efficient code, as my real dataframe has about 150000 rows and 200 columns. This gives an error:

我需要一个高效的代码，因为我真正的dataframe有大约15万行和200列。这给了一个错误:

t.d$V4<-t.d$V1[t.d$V1>1]+ t.d$V3[t.d$V3>9]

Is "apply" an option? I tried this:

“应用”是一个选择吗?我试着这样的:

t.d<-as.data.frame(matrix(1:9,ncol=3))
t.d$V4<-rep(0,nrow(t.d))

my.fun<-function(x,y){
  if(x>1 && y<9){
    x+y}
}

t.d$V4<-apply(X=t.d,MAR=1,FUN=my.fun,x=t.d$V1,y=t.d$V3)

but it gives an error as well. Thanks very much for your help.

但它也会产生错误。非常感谢你的帮助。

3 个解决方案

#1

This operation doesn't require loops, apply statements or if statements. Vectorised operations and subsetting is all you need:

此操作不需要循环、应用语句或if语句。矢量化操作和子设置是您所需要的:

t.d <- within(t.d, V4 <- V1 + V3)
t.d[!(t.d$V1>1 & t.d$V3<9), "V4"] <- 0
t.d

  V1 V2 V3 V4
1  1  4  7  0
2  2  5  8 10
3  3  6  9  0

Why does this work?

为什么这个工作吗?

In the first step I create a new column that is the straight sum of columns V1 and V4. I use within as a convenient way of referring to the columns of d.f without having to write d.f$V all the time.

在第一步中，我创建了一个新的列，该列是V1和V4的直接和。我用in来表示d的列。f，不用写d。f $ V。

In the second step I subset all of the rows that don't fulfill your conditions and set V4 for these to 0.

在第二步中，我将所有不满足条件的行集中在一起，将V4设置为0。

#2

ifelse is your friend here:

如果你是这里的朋友:

t.d$V4<-ifelse((t.d$V1>1)&(t.d$V3<9), t.d$V1+ t.d$V3, 0)

#3

I'll chip in and provide yet another version. Since you want zero if the condition doesn't mach, and TRUE/FALSE are glorified versions of 1/0, simply multiplying by the condition also works:

我将插入并提供另一个版本。如果条件不马赫，你想要零，而真/假则是1/0的美化版本，简单地乘以条件也可以:

t.d<-as.data.frame(matrix(1:9,ncol=3))
t.d <- within(t.d, V4 <- (V1+V3)*(V1>1 & V3<9))

...and it happens to be faster than the other solutions ;-)

…而且它恰好比其他的解快;

t.d <- data.frame(V1=runif(2e7, 1, 2), V2=1:2e7, V3=runif(2e7, 5, 10))
system.time( within(t.d, V4 <- (V1+V3)*(V1>1 & V3<9)) )         # 3.06 seconds
system.time( ifelse((t.d$V1>1)&(t.d$V3<9), t.d$V1+ t.d$V3, 0) ) # 5.08 seconds
system.time( { t.d <- within(t.d, V4 <- V1 + V3); 
               t.d[!(t.d$V1>1 & t.d$V3<9), "V4"] <- 0 } )       # 4.50 seconds

#1