根据行折叠数据帧并计算加权均值r
[英]collapse dataframe based on rows and calculate weighted mean r
问题描述
I would like to collapse the following dataframe 
我想折叠以下数据框
df
Chromosome Start End lengthMB imba log2 Cn mCn Cn_
chr1 0 8022945 8.023 0.026905119 -0.001671481 2 1 1.99
chr1 8022945 9168284 1.145 0.030441784 0.000601976 2 1 2
chr1 9168284 9598904 0.431 NA -0.024952441 2 1 1.91
chr1 9598904 31392788 21.794 0.036011994 0.002151497 3 1 3.01
chr2 0 8022930 8.023 0.026905119 -0.001671481 3 1 2.89
chr2 8022930 9168284 1.145 0.030441784 0.000601976 2 1 1.87
chr2 9168284 9598904 0.431 NA -0.024952441 2 1 1.57
chr2 9598904 31392788 21.794 0.036011994 0.002151497 2 0 1.87
chr2 31392788 35402000 1.164 0.029733771 0.003149921 2 1 2.01
chr3 0 8040000 1.479 NA 0.000969256 2 1 2
chr3 8040000 9168284 8.185 0.033499045 -0.031338811 1 0 0.89
chr3 9168284 9598904 3.952 0.036792754 0.002847936 1 0 0.78
chr3 9598904 31392788 0.883 0.049003807 -0.021413391 2 1 1.92
chr3 31392788 35402000 4.095 0.037653564 0.011944688 2 1 2.04
chr4 0 8022930 11.065 0.035092332 -0.022844471 2 1 1.91
chr4 8022930 9168284 40.635 0.037690844 0.006703603 2 1 2.02
chr4 9168284 9598904 0.545 0.047435696 -0.021068024 2 1 1.92
By matching only consecutive rows that have the same Cn and mCn values I want to collapse the rows. 通过仅匹配具有相同Cn和mCn值的连续行，我希望折叠这些行。 For example for the first 4 rows, we have the following: 例如，对于前4行，我们有以下内容：
Chromosome Start End lengthMB imba log2 Cn mCn Cn_
chr1 0 8022945 8.023 0.026905119 -0.001671481 2 1 1.99
chr1 8022945 9168284 1.145 0.030441784 0.000601976 2 1 2
chr1 9168284 9598904 0.431 NA -0.024952441 2 1 1.91
chr1 9598904 31392788 21.794 0.036011994 0.002151497 3 1 3.01
I want to collapse consecutive rows where they have the same Cn and mCn score, so for the first three rows that each has a "2" and "1" on the Cn and mCn column respectively and also to change the End column to reflect this collapse. 我想折叠具有相同Cn和mCn分数的连续行，因此对于前三行，在Cn和mCn列上分别分别具有“ 2”和“ 1”，并更改End列以反映这一点坍方。
Chromosome Start End lengthMB imba log2 Cn mCn Cn_
chr1 0 9598904 8.023 0.026905119 -0.001671481 2 1 1.99
But I would also like to change the Cn_column so that it is the weighted average Cn_dependant on what the lengthMB score is for that row. 但我也想更改Cn_column ，使其成为加权平均值Cn_dependant ，具体lengthMB该行的lengthMB得分。 So for the first three rows the calculation would be: 因此，对于前三行，计算将为：
((8.023/9.599) * 1.99) + ((1.145/9.599) * 2) + ((0.431/9.599) * 1.91) = 1.987
output for the first four unique Chromosomes: 前四个独特染色体的输出：
Chromosome Start End lengthMB imba log2 Cn mCn Cn_
chr1 0 9598904 8.023 0.026905119 -0.001671481 2 1 1.99
chr1 9598904 31392788 21.794 0.036011994 0.002151497 3 1 3.01
chr2 0 8022930 8.023 0.026905119 -0.001671481 3 1 2.89
chr2 8022930 9598904 1.145 0.030441784 0.000601976 2 1 1.79
chr2 9598904 31392788 21.794 0.036011994 0.002151497 2 0 1.87
chr2 31392788 35402000 1.164 0.029733771 0.003149921 2 1 2.01
chr3 0 8040000 1.479 NA 0.000969256 2 1 2
chr3 8040000 9598904 8.185 0.033499045 -0.031338811 1 0 0.836
chr3 9598904 35402000 0.883 0.049003807 -0.021413391 2 1 2.02
chr4 0 9598904 11.065 0.035092332 -0.022844471 2 1 2
Tried something like this, but I also don't know how to include the calculation... 尝试过这样的事情，但我也不知道如何包括计算...
squish_segments <- function(sample) {
setDT(sample)[, .ind:= cumsum(c(TRUE,Start[-1]!=End[-.N])),
list(lengthMB, probes, snps, imba, log2, Cn, mCn, Cn_)][,
list(Chr=Chromosome[1], Start=Start[1], End=End[.N]),
list(lengthMB, probes, snps, imba, log2, Cn, mCn, Cn_, .ind)][,.ind:=NULL][]
}
4 个解决方案
解决方案1
1 2015-11-03 13:17:35
First, please make your questions more reproducible by providing the dput output of your dataset. 首先，请通过提供数据集的dput输出使您的问题更容易重现。
I think this is what you want at a low level. 我认为这是您所需要的。
setkey(df, Chromosome, Cn, mCn, Start)
df[, list(
Start=min(Start),
End=max(End),
lengthMB=lengthMB[1],
imba=imba[1],
log2=log2[1],
Cn_=weighted.mean(Cn_, lengthMB)
), keyby=list(Chromosome, Cn , mCn)]
解决方案2
1 2015-11-03 13:28:40
This is a dplyr approach. 这是dplyr方法。
library(dplyr)
df = read.table(text=
"Chromosome Start End lengthMB imba log2 Cn mCn Cn_
chr1 0 8022945 8.023 0.026905119 -0.001671481 2 1 1.99
chr1 8022945 9168284 1.145 0.030441784 0.000601976 2 1 2
chr1 9168284 9598904 0.431 NA -0.024952441 2 1 1.91
chr1 9598904 31392788 21.794 0.036011994 0.002151497 3 1 3.01
chr2 0 8022930 8.023 0.026905119 -0.001671481 3 1 2.89
chr2 8022930 9168284 1.145 0.030441784 0.000601976 2 1 1.87
chr2 9168284 9598904 0.431 NA -0.024952441 2 1 1.57
chr2 9598904 31392788 21.794 0.036011994 0.002151497 2 0 1.87
chr2 31392788 35402000 1.164 0.029733771 0.003149921 2 1 2.01
chr3 0 8040000 1.479 NA 0.000969256 2 1 2
chr3 8040000 9168284 8.185 0.033499045 -0.031338811 1 0 0.89
chr3 9168284 9598904 3.952 0.036792754 0.002847936 1 0 0.78
chr3 9598904 31392788 0.883 0.049003807 -0.021413391 2 1 1.92
chr3 31392788 35402000 4.095 0.037653564 0.011944688 2 1 2.04
chr4 0 8022930 11.065 0.035092332 -0.022844471 2 1 1.91
chr4 8022930 9168284 40.635 0.037690844 0.006703603 2 1 2.02
chr4 9168284 9598904 0.545 0.047435696 -0.021068024 2 1 1.92", header=T)
df %>%
mutate(Consec = ifelse(Chromosome == dplyr::lag(Chromosome, default = Chromosome[1]) & ## flag consecutive matching chromosomes
Cn == dplyr::lag(Cn, default = Cn[1]) &
mCn == dplyr::lag(mCn, default = mCn[1]), 0, 1),
Consec = cumsum(Consec)) %>% ## create an id for consecutive matching chromosomes
group_by(Chromosome, Cn, mCn, Consec) %>%
summarize(Cn_ = sum(lengthMB * Cn_)/sum(lengthMB),
Start = min(Start),
End = max(End),
lengthMB = first(lengthMB),
imba= first(imba),
log2= first(log2)) %>%
ungroup() %>% ## only if you want to ungroup
select(Chromosome,Start,End, lengthMB,imba,log2,Cn,mCn,Cn_) %>% ## to re arrange column order
arrange(Chromosome, Start)
# Chromosome Start End lengthMB imba log2 Cn mCn Cn_
# (fctr) (int) (int) (dbl) (dbl) (dbl) (int) (int) (dbl)
# 1 chr1 0 9598904 8.023 0.02690512 -0.001671481 2 1 1.9876008
# 2 chr1 9598904 31392788 21.794 0.03601199 0.002151497 3 1 3.0100000
# 3 chr2 0 8022930 8.023 0.02690512 -0.001671481 3 1 2.8900000
# 4 chr2 8022930 9598904 1.145 0.03044178 0.000601976 2 1 1.7879569
# 5 chr2 9598904 31392788 21.794 0.03601199 0.002151497 2 0 1.8700000
# 6 chr2 31392788 35402000 1.164 0.02973377 0.003149921 2 1 2.0100000
# 7 chr3 0 8040000 1.479 NA 0.000969256 2 1 2.0000000
# 8 chr3 8040000 9598904 8.185 0.03349904 -0.031338811 1 0 0.8541823
# 9 chr3 9598904 35402000 0.883 0.04900381 -0.021413391 2 1 2.0187143
# 10 chr4 0 9598904 11.065 0.03509233 -0.022844471 2 1 1.9956599
Note that lag is a dplyr function, but also a stats package function. 请注意， lag是dplyr函数，也是stats包函数。 I have to write dplyr::lag otherwise there's a conflict when I try to specify default = within lag . 我必须写dplyr::lag否则当我尝试在lag内指定default =时会发生冲突。 I don't know if you or anyone else can replicate this issue. 我不知道您或其他任何人都可以复制此问题。
解决方案3
0 2015-11-03 13:19:24
Could identify unique "events" (consecutive rows with same Cn and mCn score) and then simply loop through those events and modify the rows accordingly. 可以识别唯一的“事件”（具有相同Cn和mCn分数的连续行），然后简单地遍历这些事件并相应地修改行。 Not the most efficient but should do the job. 不是最有效的，但是应该做的。
txt <- "Chromosome Start End lengthMB imba log2 Cn mCn Cn_
chr1 8022945 9168284 1.145 0.030441784 0.000601976 2 1 2
chr1 9168284 9598904 0.431 NA -0.024952441 2 1 1.91
chr1 9598904 31392788 21.794 0.036011994 0.002151497 3 1 3.01
chr2 0 8022930 8.023 0.026905119 -0.001671481 3 1 2.89
chr2 8022930 9168284 1.145 0.030441784 0.000601976 2 1 1.87
chr2 9168284 9598904 0.431 NA -0.024952441 2 1 1.57
chr2 9598904 31392788 21.794 0.036011994 0.002151497 2 0 1.87
chr2 31392788 35402000 1.164 0.029733771 0.003149921 2 1 2.01
chr3 0 8040000 1.479 NA 0.000969256 2 1 2
chr3 8040000 9168284 8.185 0.033499045 -0.031338811 1 0 0.89
chr3 9168284 9598904 3.952 0.036792754 0.002847936 1 0 0.78
chr3 9598904 31392788 0.883 0.049003807 -0.021413391 2 1 1.92
chr3 31392788 35402000 4.095 0.037653564 0.011944688 2 1 2.04
chr4 0 8022930 11.065 0.035092332 -0.022844471 2 1 1.91
chr4 8022930 9168284 40.635 0.037690844 0.006703603 2 1 2.02
chr4 9168284 9598904 0.545 0.047435696 -0.021068024 2 1 1.92"
df <- read.table(text=txt, header=T)
#identify each unique event
df$eventid <- with(df, cumsum(c(1,diff(as.numeric(factor(Chromosome)))!=0 | diff(Cn)!=0 | diff(mCn)!=0)))
#loop through events
for(i in 1:max(df$eventid)){
#identify rows in df with ith event
rows.i <- which(df$eventid == i)
df[rows.i,] <- within(df[rows.i,],{
#calculate values of interest and assign to first row of event
Start[1] <- min(Start)
End[1] <- max(End)
Cn_[1] <- sum((lengthMB/sum(lengthMB))*Cn_)
lengthMB[1] <- sum(lengthMB)
})
#drop all but first row
if(length(rows.i) > 1) df <- df[-rows.i[-1],]
} #end i
result 结果
> df
Chromosome Start End lengthMB imba log2 Cn mCn Cn_ eventid
1 chr1 8022945 9598904 1.576 0.03044178 0.000601976 2 1 1.9753871 1
3 chr1 9598904 31392788 21.794 0.03601199 0.002151497 3 1 3.0100000 2
4 chr2 0 8022930 8.023 0.02690512 -0.001671481 3 1 2.8900000 3
5 chr2 8022930 9598904 1.576 0.03044178 0.000601976 2 1 1.7879569 4
7 chr2 9598904 31392788 21.794 0.03601199 0.002151497 2 0 1.8700000 5
8 chr2 31392788 35402000 1.164 0.02973377 0.003149921 2 1 2.0100000 6
9 chr3 0 8040000 1.479 NA 0.000969256 2 1 2.0000000 7
10 chr3 8040000 9598904 12.137 0.03349904 -0.031338811 1 0 0.8541823 8
12 chr3 9598904 35402000 4.978 0.04900381 -0.021413391 2 1 2.0187143 9
14 chr4 0 9598904 52.245 0.03509233 -0.022844471 2 1 1.9956599 10
解决方案4
0 2015-11-03 16:06:16
If I understood your question properly, you can do it in one line with data.table fast grouping. 如果我正确理解了您的问题，则可以使用data.table快速分组将其data.table一行。
library(data.table)
dt[, Cn_dependent := sum((lengthMB/sum(lengthMB)) * Cn_),
by = .(Chromosome, Cn, mCn)]
to get this: 得到这个：
> dt
Chromosome Start End lengthMB imba log2 Cn mCn Cn_ Cn_dependent
1: chr1 0 8022945 8.023 0.02690512 -0.001671481 2 1 1.99 1.987601
2: chr1 8022945 9168284 1.145 0.03044178 0.000601976 2 1 2.00 1.987601
3: chr1 9168284 9598904 0.431 NA -0.024952441 2 1 1.91 1.987601
4: chr1 9598904 31392788 21.794 0.03601199 0.002151497 3 1 3.01 3.010000
5: chr2 0 8022930 8.023 0.02690512 -0.001671481 3 1 2.89 2.890000
6: chr2 8022930 9168284 1.145 0.03044178 0.000601976 2 1 1.87 1.882285
7: chr2 9168284 9598904 0.431 NA -0.024952441 2 1 1.57 1.882285
8: chr2 9598904 31392788 21.794 0.03601199 0.002151497 2 0 1.87 1.870000
9: chr2 31392788 35402000 1.164 0.02973377 0.003149921 2 1 2.01 1.882285
To collapse by Chromosome , Cn and mCn , you could use keys and unique . 要通过Chromosome ， Cn和mCn ，可以使用键和unique键。
> setkey(dt, "Chromosome", "Cn", "mCn")
> unique(dt)
Chromosome Start End lengthMB imba log2 Cn mCn Cn_ Cn_dependent
1: chr1 0 8022945 8.023 0.02690512 -0.001671481 2 1 1.99 1.987601
2: chr1 9598904 31392788 21.794 0.03601199 0.002151497 3 1 3.01 3.010000
3: chr2 9598904 31392788 21.794 0.03601199 0.002151497 2 0 1.87 1.870000
4: chr2 8022930 9168284 1.145 0.03044178 0.000601976 2 1 1.87 1.882285
5: chr2 0 8022930 8.023 0.02690512 -0.001671481 3 1 2.89 2.890000
Here is a dput of the data.table I started with: 这里是一个dput中的data.table我开始：
> dput(dt)
structure(list(Chromosome = structure(c(1L, 1L, 1L, 1L, 2L, 2L,
2L, 2L, 2L), .Label = c("chr1", "chr2"), class = "factor"), Start = c(0L,
8022945L, 9168284L, 9598904L, 0L, 8022930L, 9168284L, 9598904L,
31392788L), End = c(8022945L, 9168284L, 9598904L, 31392788L,
8022930L, 9168284L, 9598904L, 31392788L, 35402000L), lengthMB = c(8.023,
1.145, 0.431, 21.794, 8.023, 1.145, 0.431, 21.794, 1.164), imba = c(0.026905119,
0.030441784, NA, 0.036011994, 0.026905119, 0.030441784, NA, 0.036011994,
0.029733771), log2 = c(-0.001671481, 0.000601976, -0.024952441,
0.002151497, -0.001671481, 0.000601976, -0.024952441, 0.002151497,
0.003149921), Cn = c(2L, 2L, 2L, 3L, 3L, 2L, 2L, 2L, 2L), mCn = c(1L,
1L, 1L, 1L, 1L, 1L, 1L, 0L, 1L), Cn_ = c(1.99, 2, 1.91, 3.01,
2.89, 1.87, 1.57, 1.87, 2.01)), .Names = c("Chromosome", "Start",
"End", "lengthMB", "imba", "log2", "Cn", "mCn", "Cn_"), class = c("data.table",
"data.frame"), row.names = c(NA, -9L), .internal.selfref = <pointer: 0x26abf68>)
声明:本站的技术帖子网页，遵循CC BY-SA 4.0协议，如果您需要转载，请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.
相关问题
计算R数据帧中的加权平均值 - Calculate weighted average in R dataframe
计算R中的移动新近加权平均值 - Calculate moving recency-weighted mean in R
从R中的矩阵计算加权平均值 - Calculate weighted mean from matrix in R
dataframe 中加权平均值的引导跨行 - Bootstrap of weighted mean in dataframe across rows
计算数据框中几列的平均值（和加权平均值） - Calculate the mean (and weighted mean) of several columns within a dataframe
在 r 中按组计算加权滚动平均值？ - Calculate a Weighted Rolling Average by rows by group in r?
R中基于几行的加权平均值 - Weighted average in R based on a few rows
使用列名称作为权重以计算R中的加权平均值 - Use column name as weight to calculate weighted mean in R
计算R中加权离散随机变量的均值和方差 - calculate mean and variance for weighted discrete random variables in R
如何在R中使用mutate_at计算加权平均值？ - How to calculate weighted mean using mutate_at in R?
相关标签