I conducted t test on two data frames and stored results in two separate variables. They turned out to be lists. Now, I want to make a dataframe of t scores and p-values but I am not sure how to do that. I guess the lists are s3 class. The code.

```
AML_ttest <- apply(aml_df,1,t.test)
nrml_ttest <- apply(nrml_df,1,t.test)
```

Running `AML_ttest[[9]]`

gives the following result.

```
One Sample t-test
data: newX[, i]
t = 25.994, df = 25, p-value < 2.2e-16
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
6.612063 7.749997
sample estimates:
mean of x
7.18103
```

How I can get the t and p-value from each list element? And make a new dataframe?

Thanks.

### —UPDATE—

I tried the following code.

```
# AML
AML_ttest <- apply(aml_df,1,t.test)
AML_ttest = do.call(rbind,AML_ttest)
res_AML_ttest <- AML_ttest[,c("statistic","p.value")]
# Normal
nrml_ttest <- apply(nrml_df,1,t.test)
nrml_ttest = do.call(rbind,nrml_ttest)
res_nrml_ttest <- nrml_ttest[,c("statistic","p.value")]
# Make df
df_ttest <- data.frame(res_AML_ttest, res_nrml_ttest)
df_ttest
# Output
statistic p.value statistic.1 p.value.1
1 56.71269 6.171562e-28 144.5161 1.569932e-52
2 75.79649 4.559861e-31 74.87025 5.317292e-42
3 17.68306 1.207297e-15 15.15478 1.891711e-17
4 108.4904 5.984139e-35 168.8557 4.993433e-55
5 152.8165 1.156183e-38 192.4672 3.959361e-57
6 63.21714 4.163004e-29 90.42468 5.112986e-45
```

Is this approach good? AM I good to go?

### >Solution :

I’ll test with this sample data:

```
TT1 <- apply(mtcars[1:3], 2, t.test)
```

We can peek at the `str`

ucture of one of them to see what names to look for.

```
str(TT1[[1]])
# List of 10
# $ statistic : Named num 18.9
# ..- attr(*, "names")= chr "t"
# $ parameter : Named num 31
# ..- attr(*, "names")= chr "df"
# $ p.value : num 1.53e-18
# $ conf.int : num [1:2] 17.9 22.3
# ..- attr(*, "conf.level")= num 0.95
# $ estimate : Named num 20.1
# ..- attr(*, "names")= chr "mean of x"
# $ null.value : Named num 0
# ..- attr(*, "names")= chr "mean"
# $ stderr : num 1.07
# $ alternative: chr "two.sided"
# $ method : chr "One Sample t-test"
# $ data.name : chr "newX[, i]"
# - attr(*, "class")= chr "htest"
```

This suggests we can use `"statistic"`

and `"p.value"`

for that. A direct approach:

```
TT1[[1]][ c("statistic", "p.value") ]
# $statistic
# t
# 18.85693
# $p.value
# [1] 1.526151e-18
```

We can gather them together with something like:

```
out <- do.call(rbind.data.frame, lapply(TT1, `[`, c("statistic", "p.value")))
out
# statistic p.value
# mpg 18.85693 1.526151e-18
# cyl 19.59872 5.048147e-19
# disp 10.53069 9.189065e-12
```

where `mpg`

and such are row-names. They can be brought in to the frame itself if needed,

```
out$name <- rownames(out)
rownames(out) <- NULL # aesthetics only, not required
out
# statistic p.value name
# 1 18.85693 1.526151e-18 mpg
# 2 19.59872 5.048147e-19 cyl
# 3 10.53069 9.189065e-12 disp
```

(Since you made your list of t-tests with `apply(MARGIN=1)`

, though, it may not be named meaningfully if the original data did not have good row names.)