线性模型和dplyr - 更好的解决方案？ [英] Linear model and dplyr - a better solution?
问题描述
我在我最近问的问题中收到了很多好的反馈并引导使用dplyr来转换一些数据。我有一个lm（）的问题，并尝试从这个转换的数据找到一个斜率，并认为我会打开一个新的问题。
首先我有数据看起来像这样：
Var1 Var2 Var3时间温度
awj 9/9/2014 20
awj 9 / 9/2014 15
awk 9/20/2014 10
awj 9/10/2014 0
bx L 9/12/2014 30
bx L 9/12/2014 10
byk 9/13/2014 20
byk 9/13/2014 15
czj 9/14/2014 20
czj 9/14/2014 10
czk 9/14/2014 11
cwl 9/10/2014 45
adj 9/22/2014 20
adk 9/15/2014 4
adl 9/15/2014 23
adk 9/15/2014 11
我希望以此形式（Slope和Pearson模拟图示）：
V1 V2 V3斜坡皮尔逊
awj -3 -0.9
awk 2 0
adj 1.5 0.6
adk 0 0.5
adl -0.5 -0.6
bx L 12 0.7
byk 4 0.6
czj -1 -0.5
czk -3 -0.4
cwl -10 -0.9
斜率为线性 - 最小二乘法斜率。在理论上，脚本将如下所示：
library（dplyr）
data< read.table（clipboard，sep =\t，quote =，header = T）
newdata = summarize（group_by（data
，Var1
，Var2
，Var3
）
，Slope = lm（Temp〜Time）$ coeff [2]
，Pearson = cor（Time，Temp，method =pearson）
）
但是R抛出一个错误，因为它找不到Time或Temp。它可以运行 lm（data $ Temp〜data $ Time）$ coeff [2] ，但返回整个数据集的斜率，而不是我的子集寻找。 cor（）似乎在 group_by 部分中运行正常，所以我需要传递一个特定的语法 lm（）以类似的方式运行或完全使用不同的函数来获取从子集传递的斜率？
这里有几个问题。
- 如果您将数据分为3个变量（甚至是2个），则您没有足够的不同值才能运行线性回归模型
- Pearson需要两个数字值，而
Time是一个转换为数字的因素不会很有意义 - 这里的第三个问题是您需要使用
do才能运行线性模型
以下是仅在 V1上分组的图示
data％>％/ pre>
group_by（Var1）％>％＃如果您的真实数据集启用，您可以添加其他分组变量
do（mod = lm（Temp 〜％）$％
mutate（Slope = summary（mod）$ coeff [2]）％>％
select（-mod）
＃本地数据帧[3 x 2]
＃组：< by row>
＃
＃Var1 Slope
＃1 a 12.66667
＃2 b -2.50000
＃3 c -31.33333
如果您有两个数字变量，可以使用
do为了计算相关性，例如（我将创建一些虚拟数字变量来说明）data％> ;％
mutate（test1 = sample（1：3，n（），replace = TRUE），＃创建一些数值变量
test2 = sample（1：3，n（），replace = TRUE） ）％>％
group_by（Var1）％>％
do（mod = lm（Temp〜Time，data =。），
mod2 = cor（。$ test1， test2，method =pearson））％>％
mutate（Slope = summary（mod）$ coeff [2]，
Pearson = mod2 [1]）％>％
选择（-mod，-mod2）
＃来源：本地数据框[3 x 3]
＃组：< by row>
＃
＃Var1斜率皮尔森
＃1 a 12.66667 0.25264558
＃2 b -2.50000 -0.09090909
＃3 c -31.33333 0.30151134
奖金解决方案：您可以使用
data.tablepackage toolibrary（data.table）
setDT（data ）[，list（Slope = summary（lm（Temp〜Time））$ coeff [2]），Var1]
＃Var1 Slope
＃1：a 12.66667
＃2：b - 2.50000
＃3：c -31.33333
或者如果我们还要创建一些虚拟变量
library（data.table）
setDT（data）[，`：=`（test1 = sample ：3，.N，replace = TRUE），
test2 = sample（1：3，.N，replace = TRUE））] [，
list（Slope = summary（lm（Temp〜Time） ）$ coeff [2]，
Pearson = cor（test1，test2，method =pearson）），Var1]
＃Var1 Slope Pearson
＃1：a 12 .66667 -0.02159168
＃2：b -2.50000 -0.81649658
＃3：c -31.33333 -1.00000000
I got a lot of good feedback on a question I recently asked and was guided to use dplyr to transform some data. I'm having an issue with lm() and trying to find a slope from this transformed data and thought I'd open up a new question.
First I have data that looks like this:
Var1 Var2 Var3 Time Temp a w j 9/9/2014 20 a w j 9/9/2014 15 a w k 9/20/2014 10 a w j 9/10/2014 0 b x L 9/12/2014 30 b x L 9/12/2014 10 b y k 9/13/2014 20 b y k 9/13/2014 15 c z j 9/14/2014 20 c z j 9/14/2014 10 c z k 9/14/2014 11 c w l 9/10/2014 45 a d j 9/22/2014 20 a d k 9/15/2014 4 a d l 9/15/2014 23 a d k 9/15/2014 11And I want it in the form of this (values for Slope and Pearson simulated for illustration):
V1 V2 V3 Slope Pearson a w j -3 -0.9 a w k 2 0 a d j 1.5 0.6 a d k 0 0.5 a d l -0.5 -0.6 b x L 12 0.7 b y k 4 0.6 c z j -1 -0.5 c z k -3 -0.4 c w l -10 -0.9The slope being a linear-least-squares slope. In theory, the script would look like so:
library(dplyr) data <- read.table("clipboard",sep="\t",quote="",header=T) newdata = summarise(group_by(data ,Var1 ,Var2 ,Var3 ) ,Slope = lm(Temp ~ Time)$coeff[2] ,Pearson = cor(Time, Temp, method="pearson") )But R throws an error like it can't find Time or Temp. It can run
lm(data$Temp ~ data$Time)$coeff[2], but returns the slope for the entire data set and not the subsetted form that I'm looking for.cor()seems to run just fine in thegroup_bysection, so is there a specific syntax I need to pass tolm()to have it run in a similar manner or use a different function entirely to get a slope passed from the subset?解决方案You have several issues here.
- If you group your data by 3 variables (or even 2) you don't have enough distinct values in order to run a linear regression model
- Pearson requires two numeric values, while
Timeis a factor which converting to numeric won't make much sense- The third issue here is that you will need to use
doin order to run your linear modelHere's an illustration for grouping only on
V1data %>% group_by(Var1) %>% # You can add here additional grouping variables if your real data set enables it do(mod = lm(Temp ~ Time, data = .)) %>% mutate(Slope = summary(mod)$coeff[2]) %>% select(-mod) # Source: local data frame [3 x 2] # Groups: <by row> # # Var1 Slope # 1 a 12.66667 # 2 b -2.50000 # 3 c -31.33333
If you do have two numeric variables, you can use
doin order to calculate correlation too, for example (I will create some dummy numeric variables for illustration)data %>% mutate(test1 = sample(1:3, n(), replace = TRUE), # Creating some numeric variables test2 = sample(1:3, n(), replace = TRUE)) %>% group_by(Var1) %>% do(mod = lm(Temp ~ Time, data = .), mod2 = cor(.$test1, .$test2, method = "pearson")) %>% mutate(Slope = summary(mod)$coeff[2], Pearson = mod2[1]) %>% select(-mod, -mod2) # Source: local data frame [3 x 3] # Groups: <by row> # # Var1 Slope Pearson # 1 a 12.66667 0.25264558 # 2 b -2.50000 -0.09090909 # 3 c -31.33333 0.30151134
Bonus solution: you can do this quite efficiently/easily with
data.tablepackage toolibrary(data.table) setDT(data)[, list(Slope = summary(lm(Temp ~ Time))$coeff[2]), Var1] # Var1 Slope # 1: a 12.66667 # 2: b -2.50000 # 3: c -31.33333Or if we want to create some dummy variables too
library(data.table) setDT(data)[, `:=`(test1 = sample(1:3, .N, replace = TRUE), test2 = sample(1:3, .N, replace = TRUE))][, list(Slope = summary(lm(Temp ~ Time))$coeff[2], Pearson = cor(test1, test2, method = "pearson")), Var1] # Var1 Slope Pearson # 1: a 12.66667 -0.02159168 # 2: b -2.50000 -0.81649658 # 3: c -31.33333 -1.00000000
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