My data from excel file were arranged by horizontal, I would like to test its normality by row. But it will return an error if I do like this.
data<- wprov
shapiro.test(as.numeric(data[1:30]))
It will work on 1 row like this:
shapiro.test(as.numeric(data[6,c(11:40)]))
How can I make this code to compute shapiro by row?
When starting with an example data.frame like this:
(iris_head <- head(iris))
#> Sepal.Length Sepal.Width Petal.Length Petal.Width Species
#> 1 5.1 3.5 1.4 0.2 setosa
#> 2 4.9 3.0 1.4 0.2 setosa
#> 3 4.7 3.2 1.3 0.2 setosa
#> 4 4.6 3.1 1.5 0.2 setosa
#> 5 5.0 3.6 1.4 0.2 setosa
#> 6 5.4 3.9 1.7 0.4 setosa
, you could iterate over rows of the relevant subset with apply() and get a list of htest objects, one for each row:
shap_lst <- apply(iris_head[1:4], MARGIN = 1, shapiro.test)
str(shap_lst)
#> List of 6
#> $ 1:List of 4
#> ..$ statistic: Named num 0.969
#> .. ..- attr(*, "names")= chr "W"
#> ..$ p.value : num 0.838
#> ..$ method : chr "Shapiro-Wilk normality test"
#> ..$ data.name: chr "newX[, i]"
#> ..- attr(*, "class")= chr "htest"
#> $ 2:List of 4
#> ..$ statistic: Named num 0.982
#> .. ..- attr(*, "names")= chr "W"
#> ..$ p.value : num 0.914
#> ..$ method : chr "Shapiro-Wilk normality test"
#> ..$ data.name: chr "newX[, i]"
#> ..- attr(*, "class")= chr "htest"
#> $ 3:List of 4
#> ..$ statistic: Named num 0.971
#> .. ..- attr(*, "names")= chr "W"
#> ..$ p.value : num 0.846
#> ..$ method : chr "Shapiro-Wilk normality test"
#> ..$ data.name: chr "newX[, i]"
#> ..- attr(*, "class")= chr "htest"
#> $ 4:List of 4
#> ..$ statistic: Named num 0.987
#> .. ..- attr(*, "names")= chr "W"
#> ..$ p.value : num 0.94
#> ..$ method : chr "Shapiro-Wilk normality test"
#> ..$ data.name: chr "newX[, i]"
#> ..- attr(*, "class")= chr "htest"
#> $ 5:List of 4
#> ..$ statistic: Named num 0.963
#> .. ..- attr(*, "names")= chr "W"
#> ..$ p.value : num 0.796
#> ..$ method : chr "Shapiro-Wilk normality test"
#> ..$ data.name: chr "newX[, i]"
#> ..- attr(*, "class")= chr "htest"
#> $ 6:List of 4
#> ..$ statistic: Named num 0.968
#> .. ..- attr(*, "names")= chr "W"
#> ..$ p.value : num 0.831
#> ..$ method : chr "Shapiro-Wilk normality test"
#> ..$ data.name: chr "newX[, i]"
#> ..- attr(*, "class")= chr "htest"
Those can be converted to a frame and added to your dataset, if needed:
# drop htest class for bind_rows()
shap_df <- lapply(shap_lst, unclass) |> dplyr::bind_rows()
shap_df
#> # A tibble: 6 × 4
#> statistic p.value method data.name
#> <dbl> <dbl> <chr> <chr>
#> 1 0.969 0.838 Shapiro-Wilk normality test newX[, i]
#> 2 0.982 0.914 Shapiro-Wilk normality test newX[, i]
#> 3 0.971 0.846 Shapiro-Wilk normality test newX[, i]
#> 4 0.987 0.940 Shapiro-Wilk normality test newX[, i]
#> 5 0.963 0.796 Shapiro-Wilk normality test newX[, i]
#> 6 0.968 0.831 Shapiro-Wilk normality test newX[, i]
# add statistic & p.value columns to iris_head
cbind(iris_head, shap_df[,1:2])
#> Sepal.Length Sepal.Width Petal.Length Petal.Width Species statistic p.value
#> 1 5.1 3.5 1.4 0.2 setosa 0.9694987 0.8383766
#> 2 4.9 3.0 1.4 0.2 setosa 0.9820427 0.9138954
#> 3 4.7 3.2 1.3 0.2 setosa 0.9707126 0.8458912
#> 4 4.6 3.1 1.5 0.2 setosa 0.9867607 0.9403457
#> 5 5.0 3.6 1.4 0.2 setosa 0.9627830 0.7964093
#> 6 5.4 3.9 1.7 0.4 setosa 0.9683317 0.8311267
Thank you, let me try and do this one.
I have tried this one and it worked. However, I planned to exclude the zero in the data but when I assigned ;
data[data==0]<-NA
and perform the same code
I get an error like this.
And my actual data supposed to be arranged like this way.
I already got this one, but somehow, when I compared the result to shapiro using one specific row and the result of shapiro by row. it has come up with different result?
do you get the same problem when your data is the iris example data set ? if so can you show us how to get the same problematic result ?
