更新：data.frame输出

Question

DF

sampleName     realConc      exptname concentrate timepoints replicate    day  var
name1_0     3.877049e-05           0hr        55mM          0        b1 011311   1
name1_20    3.293085e-04           kcl        55mM         20        b1 011311   2
name1_40    3.999433e-05           kcl        55mM         40        b1 011311   3
name2_0     2.939995e-03           0hr        55mM          0        b1 011411   1
name2_20    1.212584e-02           kcl        55mM         20        b1 011411   2
name2_40    1.894434e-02           kcl        55mM         40        b1 011411   3

我想将每个realConc值除以realConc值，时间点为0，具有相等的日，复制和集中值

我正在尝试for循环，而不是太多运气，你能帮帮我吗？

for (i in 1:dim(df)[1]){

df$realConc <- df$realConc[i] / df[which(duplicated(paste(replicate,day))) & df$timepoint == 0,]$realConc[i]
}

我在想这样的事情，但显然不起作用

Answer 1

plyr是你的朋友！

library(plyr)
ddply(df, .(day, replicate, concentrate),
      transform, scaled=realConc/realConc[timepoints==0])

#   sampleName     realConc exptname concentrate timepoints replicate   day var   scaled
# 1    name1_0 3.877049e-05      0hr        55mM          0        b1 11311   1 1.000000
# 2   name1_20 3.293085e-04      kcl        55mM         20        b1 11311   2 8.493793
# 3   name1_40 3.999433e-05      kcl        55mM         40        b1 11311   3 1.031566
# 4    name2_0 2.939995e-03      0hr        55mM          0        b1 11411   1 1.000000
# 5   name2_20 1.212584e-02      kcl        55mM         20        b1 11411   2 4.124442
# 6   name2_40 1.894434e-02      kcl        55mM         40        b1 11411   3 6.443664

Answer 2

您尚未指定输出的外观，但这是执行该计算的一种方法：

首先，读入您的数据（最好使用dput()或提供一些代码来重新创建数据。

test = read.table(header=TRUE, text = "sampleName     realConc      exptname concentrate timepoints replicate    day  var
name1_0     3.877049e-05           0hr        55mM          0        b1 011311   1
name1_20    3.293085e-04           kcl        55mM         20        b1 011311   2
name1_40    3.999433e-05           kcl        55mM         40        b1 011311   3
name2_0     2.939995e-03           0hr        55mM          0        b1 011411   1
name2_20    1.212584e-02           kcl        55mM         20        b1 011411   2
name2_40    1.894434e-02           kcl        55mM         40        b1 011411   3")

然后，根据您需要的分组拆分数据。

temp = split(test, list(test$day, test$concentrate, test$replicate))

第三，按组计算realConc的{{1}}值，然后用它来进行划分。

timepoints == 0

更新：data.frame输出

lapply(temp, function(x) x[, 2]/x[which(x$timepoints == 0), 2])
# $`11311.55mM.b1`
# [1] 1.000000 8.493793 1.031566
# 
# $`11411.55mM.b1`
# [1] 1.000000 4.124442 6.443664

Answer 3

这是一个简单的基础R版本。调试可能更容易，因为每个步骤都是可见的。

# find all possible denominators and rename realConc to avoid duplicate name in merge
denom <- x[x$timepoints == 0,c('realConc','concentrate','replicate','day')]
names(denom)[1] <- 'realConcDenominator'

# merge in new column with appropriate denominator
x$realConcDenominator <- merge(x,denom,by = c('concentrate','replicate','day'),all.x = T)[,'realConcDenominator']

# and divide
x$result <- x$realConc / x$realConcDenominator

另一个使用申请。

# or use apply in one shot
x$applyresult <- apply(x,1,function(x,denom){
  as.numeric(x['realConc'])/denom[denom$concentrate == x['concentrate'] & denom$replicate == x['replicate'] & denom$day == x['day'],'realConc']
},denom = x[x$timepoints == 0,c('realConc','concentrate','replicate','day')])

循环一些简单的划分

3 个答案:

更新：data.frame输出