R-如何在另一列的条件定义的间隔之间求和一列中的对象 [英] R - How to sum objects in a column between an interval defined by conditions on another column

问题描述

这是对此问题的一种应用:

This comes as an application to this question:Sum object in a column between an interval defined by another column

对于((A[i+1]-A[i]==0)或(A[i+1]-A[i]==1)或(A[i]-A[i-1]==0)或(A[i]-A[i-1]==1))(其中i是行索引)，如果我想对B中的值求和，我想知道如何调整答案.所以基本上将A的B行总和等于+/- 1，但同一行两次不求和?

What I would like to know is how to adjust the answer if I want to sum the values in B, for ((A[i+1]-A[i]==0) or (A[i+1]-A[i]==1) or (A[i]-A[i-1]==0) or (A[i]-A[i-1]==1)) where i is the row index, so basically sum B rows for A-s that have the same value +/- 1, but not sum the same row twice?

我尝试构建循环函数，但在将行索引与数据帧一起使用时卡住了. 例子: 如果给出以下数据帧

I tried building a loop function but I get stuck when using row indices with data frames. Example: If the following data frame is given

df     
      A B
[1,]  1 4
[2,]  1 3
[3,]  3 5
[4,]  3 7
[5,]  4 3
[6,]  5 2

我要获取的是下一个数据帧:

What I want to obtain is the next data frame:

df
      A B
[1,]  1 7
[2,]  3 15
[3,]  5 2

此外，如果一个具有较大的数据帧，例如:

Moreover if a have a large data frame like this:

df
chr     start           stop            m       n       s
chr1    71533361        71533362        23      1       -
chr1    71533361        71533362        24      26      -
chr1    71533361        71533362        25      1       -

，我希望我的结果看起来像这样(我选择了m列中的值为max的行):

and I want my result to look like this (I chose the row for which the value in column m is max):

df
chr1    71533361        71533362        24      28      -

推荐答案

假设您的原始数据框为df，请尝试以下操作:

Try the following, assuming your original dataframe is df:

df2 <- df # create a duplicate df to destroy
z <- data.frame(nrow=length(unique(df$A)), ncol=2) # output dataframe
names(z) <- c("A","B")
j <- 1 # output indexing variable
u <- unique(df$A) # unique vals of A
i <- u[1]
s <- TRUE # just for the while() loop
while(s){
    z[j,] <- c(i,sum(df2[df2$A %in% c(i-1,i,i+1),2]))
    df2 <- df2[!df2$A %in% c(i-1,i,i+1),]
    j <- j + 1 # index the output
    u <- u[!u %in% c(i-1,i,i+1)] # cleanup the u vector
    if(length(u)==0) # conditionally exit the loop
        s <- FALSE
    else
        i <- min(u) # reset value to sum by
}

我知道那是一堆凌乱的代码，但是考虑到所有不同的索引，这是一个棘手的问题.

I know that's kind of messy code, but it's a sort of tough problem given all of the different indices.

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