由于字符变量，在R中运行T-Test时出现错误消息

由一场梦而已发布于 2020-05-24 07:17:13

I have been trying to run a two side t-test in R but keep running into error. Below is my process flow, dataset details and script from R-studio. I used a dataset called LungCapacity that I downloaded from this website: https://www.statslectures.com/r-scripts-datasets.

#Imported data set into RStudio.

# Ran a summary report to see the data and class.
summary(LungCapData)

# Here I could see that the smoke column is a character, so I converted it to a factor
LungCapacityData$Smoke <- factor(LungCapacityData$Smoke)

# On checking the summary. I see its converted to a factor with a yes and no.

# I want to run a t-test between lung capacity and smoking. 
t.test(LungCapData$LungCap, LungCapData$Smoke, alternative = c("two.sided"), mu=0, var.equal = FALSE, conf.level = 0.95, paired = FALSE)

现在，在运行它时，我得到以下错误。

Error in var(y) : Calling var(x) on a factor x is defunct.
  Use something like 'all(duplicated(x)[-1L])' to test for a constant vector.
In addition: Warning message:
In mean.default(y) : argument is not numeric or logical: returning NA

我试图将Smoke变量从Yes和No转换为1和0。数据正在运行，但不正确。我究竟做错了什么？

Doyle

2020-05-24 07:17:13

You're very close, you just need to call t.test with a formula:

t.test(LungCap ~ Smoke, data = LungCapacityData,
       alternative = c("two.sided"), mu=0, var.equal = FALSE,
       conf.level = 0.95, paired = FALSE)

#   Welch Two Sample t-test
#
#data:  LungCap by Smoke
#t = -3.6498, df = 117.72, p-value = 0.0003927
#alternative hypothesis: true difference in means is not equal to 0
#95 percent confidence interval:
# -1.3501778 -0.4003548
#sample estimates:
# mean in group no mean in group yes 
#         7.770188          8.645455

With your current approach, you're trying to compare LungCapData$LungCap which is a numeric vector:

LungCapData$LungCap[1:10]
# [1]  6.475 10.125  9.550 11.125  4.800  6.225  4.950  7.325  8.875  6.800

With LungCapData$Smoke, which is a vector of factors:

LungCapData$Smoke[1:10]
# [1] no  yes no  no  no  no  no  no  no  no

Instead, you want to instruct t.test to compare LungCapData$LungCap when grouping by LungCapData$Smoke. That is achieved with a formula.

The formula LungCap ~ Smoke says that LungCap should depend on Smoke. When you use a formula, you also need to supply data =.

When you try to convert LungCapData$Smoke to numeric, you get the wrong result because you're just getting the factor level indices which have no biological significance.

as.numeric(LungCapData$Smoke)[1:10]
# [1] 1 2 1 1 1 1 1 1 1 1

您基本上是在问我们分配的因子水平的平均值是否与肺活量的平均值不同。

The other way is to subset LungCapData$LungCap yourself, but that's a lot more typing:

t.test(LungCapacityData$LungCap[LungCapacityData$Smoke == "yes"],
       LungCapacityData$LungCap[LungCapacityData$Smoke == "no"],
       alternative = c("two.sided"), mu=0, var.equal = FALSE,
       conf.level = 0.95, paired = FALSE)

zamet

2020-05-24 07:17:13

As specified in the OP, t.test() attempts to compare two vectors, expecting them to be numeric.

Instead, use the formula version of t.test().

data <- read.table(file = "./data/LungCapData.txt",header = TRUE)
t.test(LungCap ~ Smoke,data = data)

...以及输出：

> t.test(LungCap ~ Smoke,data = data)

    Welch Two Sample t-test

data:  LungCap by Smoke
t = -3.6498, df = 117.72, p-value = 0.0003927
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -1.3501778 -0.4003548
sample estimates:
 mean in group no mean in group yes 
         7.770188          8.645455 

>

一场梦而已

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