如何使用 for 循环获取我的函数运行通过所有值

Question

我的函数有问题 'subtract_within'。该函数的目标是获取变量中的一个值，然后将该值与变量中所有其他值的差相加。

当我调用它时，它returns for 循环中第一个元素（正确！）的值，但不是其余元素的值。

这是我的代码：

x<- c(1,2,3,4,5,6) #the value i want to calculate with
y<- c(1,2,3,4,5,6) #the identifier of the value
this_df<- data.frame(x,y) 
this_df$z = 0 #now theres a new variable that is 0 for now but will be filled in later

this_df %>% 
  mutate(z = subtract_within(x,y)) #this is how i will ideally call on this function 'subtract_within'

#start of my function 'subtract_within'
subtract_within<- function(x,y){ #this function takes in two variables from a dataframe
  dataframe<- data.frame(x,y)  # i locally make it a dataframe
  dataframe$z<-0 # i add the variable that i will be returning
  for (i in 1:nrow(dataframe)){
    dataframe$z[i] <- (dataframe$x[i]*(length(which(dataframe$y != dataframe$y[i])))) - sum(dataframe$x[which(dataframe$y != dataframe$y[i])])
    return(dataframe$z)
  }
  return(dataframe$z)
}  
'''

My output is as follows:

 x y   z
1 1 1 -15
2 2 2   0
3 3 3   0
4 4 4   0
5 5 5   0

ideally my output would be: 

 x y   z
1 1 1 -15
2 2 2 -5
3 3 3  0
4 4 4  5
5 5 5  10

Answer 1

是在for循环中添加return的情况。如果我们只使用最后一个return，它应该可以工作

subtract_within<- function(x,y){ #this function takes in two variables from a dataframe
  dataframe<- data.frame(x,y)  # i locally make it a dataframe
  dataframe$z<-0 # i add the variable that i will be returning
  for (i in 1:nrow(dataframe)){
    dataframe$z[i] <- (dataframe$x[i]*(length(which(dataframe$y != dataframe$y[i])))) - sum(dataframe$x[which(dataframe$y != dataframe$y[i])])
    #return(dataframe$z)
  }
  return(dataframe$z)
}

Answer 2

我认为问题在于您的循环中有一个 return 语句

for (i in 1:nrow(dataframe)){
    dataframe$z[i] <- (dataframe$x[i]*(length(which(dataframe$y != dataframe$y[i])))) - sum(dataframe$x[which(dataframe$y != dataframe$y[i])])
    return(dataframe$z) # THIS RIGHT HERE
}
return(dataframe$z)

这使得函数立即return循环后的数据帧只运行一次。如果你删除它，我认为它应该有效。

for (i in 1:nrow(dataframe)){
    dataframe$z[i] <- (dataframe$x[i]*(length(which(dataframe$y != dataframe$y[i])))) - sum(dataframe$x[which(dataframe$y != dataframe$y[i])])
}
return(dataframe$z)

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