相对增长率计算--计算所有观测的增长2-增长1

Question

我试图计算出13000棵树的相对生长率，我们对其进行了9年的测量。我想计算以下公式：

RGR =(lnGrowthT2-ln GrowthT1)/(t2-t1)

因此，时间2的增长日志减去时间1的日志增长，我可以对单个树数据进行这一操作。但是，我无法在原始的data.frame中创建RGR列，增长已经用log函数(cvallog)转换了。

我的数据：

 Subset1= data[data$Id== 1|data$Id==2,]
 > Subset1
    Id          Tree replicate        cval      cvallog
1    1 A_12N_A1_ABBA      2009 0.029224666 -3.532742212
2    2 A_12N_A2_BEAL      2009 0.007306166 -4.919036573
65   1 A_12N_A1_ABBA      2010 0.317308712 -1.147880124
66   2 A_12N_A2_BEAL      2010 0.459960580 -0.776614488
134  1 A_12N_A1_ABBA      2011 0.653251467 -0.425793129
135  2 A_12N_A2_BEAL      2011 1.040621151  0.039817795
198  1 A_12N_A1_ABBA      2012 0.794725631 -0.229758343
199  2 A_12N_A2_BEAL      2012 2.042820623  0.714331511
262  1 A_12N_A1_ABBA      2013 0.735415424 -0.307319736
263  2 A_12N_A2_BEAL      2013 2.189564416  0.783702627
333  1 A_12N_A1_ABBA      2014 0.942966706 -0.058724303
334  2 A_12N_A2_BEAL      2014 3.135083588  1.142655836
397  1 A_12N_A1_ABBA      2015 0.587495498 -0.531886697
398  2 A_12N_A2_BEAL      2015 3.631681108  1.289695656
463  1 A_12N_A1_ABBA      2016 0.995382216 -0.004628479
464  2 A_12N_A2_BEAL      2016          NA           NA
527  1 A_12N_A1_ABBA      2017 0.951695307 -0.049510351
528  2 A_12N_A2_BEAL      2017 3.394292628  1.222095382

这是一棵树的代码：

Subset1= data[data$Id== 1,]  
###create the t1 and 2 t2 columns

Subset1$t1=Subset1$cvallog # create a t1 column
t2=as.data.frame((Subset1$cvallog)[-1])# remove first observation, crate a   t2 column 
t2= as.vector (rbind (t2,  0))
Subset1$t2 = t2  #add t2 to main data.frame so to  have t1 and t2 on the same row.
replicate_t2=as.data.frame((Subset1$replicate)[-1])# remove first observation, crate a t2 column 
replicate_t2= as.vector (rbind (replicate_t2,  0))
Subset1$replicate_t2 = replicate_t2

#calculate RGR  Relative growth
Subset1$RGR = ((Subset1$t2)- (Subset1$t1))/ ((Subset1$replicate_t2)-Subset1$replicate)  #difference between log values divided by 1 because we have 1 year interval

当我试图在带有拆分函数的循环中这样做时，我不能用所有的树创建一个最终的输出。

uniqTree <- (split(data, data$Tree))

for (i in 1:length(uniqTree)){
temp =data[data$Id== i,]
temp$t1=temp$cvallog #create t1 column

t2=as.data.frame((temp$cvallog)[-1]) #create t2 column
t2= as.vector (rbind (t2,  0))
temp$t2 = t2
replicate_t2=as.data.frame((temp$replicate)[-1])# remove first observation, crate a t2_replicate column 
replicate_t2= as.vector (rbind (replicate_t2,  0))
temp$replicate_t2 = replicate_t2


#calculate RGR  Relative growth
temp$RGR  = NULL
temp$RGR  = ((temp$t2)- (temp$t1))  / (temp$replicate_t2-temp$replicate)
print(temp$RGR)
}

谢谢你的帮助

Answer 1

如果使用dplyr包

df %>% arrange(Tree,replicate) %>% group_by(Tree) %>% mutate(rt = (cvallog-lag(cvallog))/(replicate-lag(replicate)))

其中df

df = read.table(text= " Id Tree replicate cval cvallog 1 1 A_12N_A1_ABBA 2009 0.029224666 -3.532742212 2 2 A_12N_A2_BEAL 2009 0.007306166 -4.919036573 65 1 A_12N_A1_ABBA 2010 0.317308712 -1.147880124 66 2 A_12N_A2_BEAL 2010 0.459960580 -0.776614488 134 1 A_12N_A1_ABBA 2011 0.653251467 -0.425793129 135 2 A_12N_A2_BEAL 2011 1.040621151 0.039817795 198 1 A_12N_A1_ABBA 2012 0.794725631 -0.229758343 199 2 A_12N_A2_BEAL 2012 2.042820623 0.714331511 262 1 A_12N_A1_ABBA 2013 0.735415424 -0.307319736 263 2 A_12N_A2_BEAL 2013 2.189564416 0.783702627 333 1 A_12N_A1_ABBA 2014 0.942966706 -0.058724303 334 2 A_12N_A2_BEAL 2014 3.135083588 1.142655836 397 1 A_12N_A1_ABBA 2015 0.587495498 -0.531886697 398 2 A_12N_A2_BEAL 2015 3.631681108 1.289695656 463 1 A_12N_A1_ABBA 2016 0.995382216 -0.004628479 464 2 A_12N_A2_BEAL 2016 NA NA 527 1 A_12N_A1_ABBA 2017 0.951695307 -0.049510351 528 2 A_12N_A2_BEAL 2017 3.394292628 1.222095382" )