Chi Square Test for Each Row in Data Frame

chi square test for each row in data frame

You can use apply with "MARGIN =1" to and then do the chisq.test. Extract the values using $statistic and $p.value and cbind it to the dataset.

 df1 <- cbind(df, t(apply(df, 1, function(x) {
             ch <- chisq.test(x)
             c(unname(ch$statistic), ch$p.value)})))

 colnames(df1)[3:4] <- c('x-squared', 'p-value')

chisq.test for each row on four numbers and output in new data frame in R

You were close, just inspect one single test with str which helps you to decide which elements to select.

apply(dat[,c('female_boxing','female_cycling','male_boxing','male_cycling')], 
      1, function(x) chisq.test(x)[c('statistic', 'p.value')] )

The apply gives you a list, the results are a little nicer using sapply and looping over the rows.

chi <- t(sapply(seq(nrow(dat)), function(i) 
  chisq.test(dat[i, c('female_boxing','female_cycling','male_boxing','male_cycling')])[
    c('statistic', 'p.value')]))

cbind(dat, chi)
#   ID1 ID2 female_boxing female_cycling male_boxing male_cycling statistic       p.value
# 1   A zit            43            170         159          710  988.7209 5.033879e-214
# 2   B tag            37            134         165          744  1142.541 2.146278e-247
# 3   C hfs            32             96         170          784  1334.991 3.762222e-289
# 4   D prt            17             61         185          811  1518.015             0
# 5   E its            31            112         169          762  1245.218 1.133143e-269
# 6   F qrw            68            233         130          645  752.3941 9.129485e-163

---

Data:

dat <- structure(list(ID1 = c("A", "B", "C", "D", "E", "F"), ID2 = c("zit", 
"tag", "hfs", "prt", "its", "qrw"), female_boxing = c(43L, 37L, 
32L, 17L, 31L, 68L), female_cycling = c(170L, 134L, 96L, 61L, 
112L, 233L), male_boxing = c(159L, 165L, 170L, 185L, 169L, 130L
), male_cycling = c(710L, 744L, 784L, 811L, 762L, 645L)), class = "data.frame", row.names = c(NA, 
-6L))

Chi-square tests for different groups in a R dataframe

base

df <- data.frame(species = factor(c(rep("species1", 4), rep("species2", 4), rep("species3", 4))),
                   trap = c(rep(c("A","B","C","D"), 3)),
                   count=c(6,3,7,9,5,3,6,6,5,8,1,3))
df
#>     species trap count
#> 1  species1    A     6
#> 2  species1    B     3
#> 3  species1    C     7
#> 4  species1    D     9
#> 5  species2    A     5
#> 6  species2    B     3
#> 7  species2    C     6
#> 8  species2    D     6
#> 9  species3    A     5
#> 10 species3    B     8
#> 11 species3    C     1
#> 12 species3    D     3

species <- unique(df$species)

chi_species <- lapply(species, function(x) xtabs(count~trap, df, 
                      subset = species== x))

chi_species <- setNames(chi_species, species)

lapply(chi_species, chisq.test)

#> $species1
#> 
#>  Chi-squared test for given probabilities
#> 
#> data:  X[[i]]
#> X-squared = 3, df = 3, p-value = 0.3916
#> 
#> 
#> $species2
#> 
#>  Chi-squared test for given probabilities
#> 
#> data:  X[[i]]
#> X-squared = 1.2, df = 3, p-value = 0.753
#> 
#> 
#> $species3
#> 
#>  Chi-squared test for given probabilities
#> 
#> data:  X[[i]]
#> X-squared = 6.2941, df = 3, p-value = 0.09815

^{Created on 2022-04-25 by the reprex package (v2.0.1)}

tidyverse

df %>% 
  group_by(species, trap) %>% 
  summarise(count = sum(count)) %>% 
  summarise(pvalue= chisq.test(count)$p.value) 

# A tibble: 3 × 2
  species  pvalue
  <fct>     <dbl>
1 species1 0.392 
2 species2 0.753 
3 species3 0.0981

add results of chi square test to each row

To see why the error is trying to communicate, compare your data with the type of data chisq.test is expecting:

dput(matrix(main[1,2:5,drop=T], nrow=2, 2,2))
# structure(list(20, 10, 40, 80), .Dim = c(2L, 2L))
dput(matrix(1:4, nrow=2, 2,2))
# structure(c(1L, 3L, 2L, 4L), .Dim = c(2L, 2L))

One remedy is to force you data into a numeric vector:

res <- chisq.test(matrix(as.numeric(main[1,2:5]), nrow=2, 2,2))
res
#   Pearson's Chi-squared test with Yates' continuity correction
# data:  matrix(as.numeric(main[1, 2:5]), nrow = 2, 2, 2)
# X-squared = 9.7656, df = 1, p-value = 0.001778

Now, if you want to add the results to each row, you first need to pick "which results". Namely, the results are actually prettied up a bit, with several tidbits internally:

str(unclass(res))
# List of 9
#  $ statistic: Named num 9.77
#   ..- attr(*, "names")= chr "X-squared"
#  $ parameter: Named int 1
#   ..- attr(*, "names")= chr "df"
#  $ p.value  : num 0.00178
#  $ method   : chr "Pearson's Chi-squared test with Yates' continuity correction"
#  $ data.name: chr "matrix(as.numeric(main[1, 2:5]), nrow = 2, 2, 2)"
#  $ observed : num [1:2, 1:2] 20 10 40 80
#  $ expected : num [1:2, 1:2] 12 18 48 72
#  $ residuals: num [1:2, 1:2] 2.309 -1.886 -1.155 0.943
#  $ stdres   : num [1:2, 1:2] 3.33 -3.33 -3.33 3.33

If you wanted to include (e.g.) the test statistic as a number, you might do:

chisq.statistic <- sapply(seq_len(nrow(main)), function(row) {
  chisq.test(matrix(as.numeric(main[row,2:5]), nrow=2, 2,2))$statistic
})
main$chisq.statistic <- chisq.statistic
main
#    Genes Group1_Mut Group1_WT Group2_Mut Group2_WT chisq.statistic
# 1 GENE_A         20        40         10        80      9.76562500
# 2 GENE_B         10        50         30        60      4.29687500
# 3 GENE_C          5        55         10        80      0.07716049

Note that tools like dplyr and data.table may facilitate this. For example:

library(dplyr)
main %>%
  rowwise() %>%
  mutate(
    chisq.statistic = chisq.test(matrix(c(Group1_Mut, Group1_WT, Group2_Mut, Group2_WT), nrow = 2))$statistic
  )
# Source: local data frame [3 x 6]
# Groups: <by row>
# # A tibble: 3 × 6
#    Genes Group1_Mut Group1_WT Group2_Mut Group2_WT chisq.statistic
#   <fctr>      <dbl>     <dbl>      <dbl>     <dbl>           <dbl>
# 1 GENE_A         20        40         10        80      9.76562500
# 2 GENE_B         10        50         30        60      4.29687500
# 3 GENE_C          5        55         10        80      0.07716049

This example shows one thing you may wish to incorporate into whichever method you use: explicit naming of columns. That is, "2:5" could change depending on your input matrix.

With R, I would like to loop through each row and create corresponding chisquare results for each row

An option is apply with MARGIN = 1 to loop over the rows. Within each row, it is a vector, so we just need to wrap with matrix to convert to a matrix with specified dimensions, apply the chisq.test and get the output in a tibble format with tidy

library(broom)
library(dplyr)
apply(df, 1, function(x) tidy(chisq.test(matrix(x, ncol = 2)))) %>%
     bind_rows

---

Or this can be done in tidyverse with pmap

library(purrr)
pmap_dfr(df, ~ c(...) %>%
             matrix(ncol = 2) %>%
             chisq.test %>%
             tidy)

-output

# A tibble: 4 x 4
#  statistic p.value parameter method                                                      
#      <dbl>   <dbl>     <int> <chr>                                                       
#1  3.17e- 1   0.574         1 Pearson's Chi-squared test with Yates' continuity correction
#2  1.66e- 2   0.898         1 Pearson's Chi-squared test with Yates' continuity correction
#3  6.70e-32   1.00          1 Pearson's Chi-squared test with Yates' continuity correction
#4  7.51e- 1   0.386         1 Pearson's Chi-squared test with Yates' continuity correction

R: How to perform a chiquare test on two groups & each row of a dataframe

First, this is a statistics question, not an R question really. You should try posting it on stats.stackexchange.com, where you are likely to get a much better answer.

Second, there are two types of chi-square test, one to assess whether a sample is from a given test distribution, and one to test for independence. I assume that you are interested in the first type.

If that is correct, then it looks like you are asking - using the first row of your df as an example - how likely is it, if allele A and allele B are present in equal amounts, that you could get a sample where allele A is present at 69% and allele B is present at 31%? If the likelihood (p) is very low, then we can assert with confidence 1-p that allele A and allele B were not present in equal amounts.

[NB: If this is not what you are asking, then I am misunderstanding your question - let me know in a comment and I'll delete the answer.]

In your case it is probably better to skip the vagaries of the chisq.test(...) function in R and go directly to the definition of Xi-sq:

χ² = Σ( O_i - E_i )² / E_i

Where O_i and E_i are the i^th observed and expected value, respectively. The way you have this set up, in each row there are only 2 observations, for allele A and allele B. So for row 1 we would write:

χ² = (0.692 - 0.5)² / 0.5 + (0.307 - 0.5)² / 0.5 = 0.148

Since there are only two observations, there is only 1 degree of freedom. Chi-square tests with only 1 df are extremely unreliable, so I do not recommend this, but following through for the sake of the example, we can calculate the probability that chi-sq will be this large or larger as follows in R:

pchisq(0.148, df=1, lower.tail=F)
# [1] 0.700454

This means, assuming that allele A and allele B are present in equal amounts, there is still a 70% chance that you could obtain a sample with allele A present at 69% and allele B present at 31%. So we definately cannot reject the null hypothesis (that allele A and B are present equally).

Running this test for all rows is straightforward:

df           <- na.omit(df)          # remove rows with missing values
colnames(df) <- c("A.obs","B.obs","A.exp","B.exp")  # because I'm lazy
df$chisq     <- with(df,(A.obs-A.exp)^2/A.exp + (B.obs-B.exp)^2/B.exp)
df$p.value   <- pchisq(df$chisq,df=1, lower.tail=F)
df
#       A.obs     B.obs A.exp B.exp    chisq   p.value
# 1 0.6923077 0.3076923   0.5   0.5 0.147929 0.7005224
# 4 0.6250000 0.3750000   0.5   0.5 0.062500 0.8025873
# 5 0.6250000 0.3750000   0.5   0.5 0.062500 0.8025873

You can in fact use the chisq.test(...) function to do this, although in your case I'm not sure it's an improvement:

t(apply(df,1,function(x)
        with(chisq.test(x[1:2],p=x[3:4]),c(statistic,p.value=p.value))))
#   X-squared   p.value
# 1  0.147929 0.7005224
# 4  0.062500 0.8025873
# 5  0.062500 0.8025873

---

Related Topics