I have SVG & it is of size 1326 x 745.
SVG looks like:
import React from 'react'
export const Wave = (props: React.SVGProps<SVGSVGElement>) => {
return (
<svg viewBox="0 0 16 9" {...props}>
<path
d="M0 8L0.2 7.8C0.3 7.7 0.7 7.3 1 7.2C1.3 7 1.7 7 2 6.8C2.3 6.7 2.7 6.3 3 6C3.3 5.7 3.7 5.3 4 5.7C4.3 6 4.7 7 5 7.5C5.3 8 5.7 8 6 7.5C6.3 7 6.7 6 7 5.7C7.3 5.3 7.7 5.7 8 5.8C8.3 6 8.7 6 9 6.2C9.3 6.3 9.7 6.7 10 6.8C10.3 7 10.7 7 11 6.7C11.3 6.3 11.7 5.7 12 5.7C12.3 5.7 12.7 6.3 13 6.5C13.3 6.7 13.7 6.3 14 6.3C14.3 6.3 14.7 6.7 15 6.8C15.3 7 15.7 7 15.8 7L16 7L16 10L15.8 10C15.7 10 15.3 10 15 10C14.7 10 14.3 10 14 10C13.7 10 13.3 10 13 10C12.7 10 12.3 10 12 10C11.7 10 11.3 10 11 10C10.7 10 10.3 10 10 10C9.7 10 9.3 10 9 10C8.7 10 8.3 10 8 10C7.7 10 7.3 10 7 10C6.7 10 6.3 10 6 10C5.7 10 5.3 10 5 10C4.7 10 4.3 10 4 10C3.7 10 3.3 10 3 10C2.7 10 2.3 10 2 10C1.7 10 1.3 10 1 10C0.7 10 0.3 10 0.2 10L0 10Z"
strokeLinecap="round"
strokeLinejoin="miter"
fill="currentColor"
></path>
</svg>
)
}
It takes so much space. I want it to be really small. Only take up wave space.
I made this Wave SVG on https://app.haikei.app/
I tried all configurations but I can't get it to be small.
All I want is width to be 100% no matter which device I'm on & it should have a proportional height. What do I need to do to change it?
Related
trying to determine how I would address this problem using R code.
Brief description of problem:
There is a minimum mass required to run an analysis of samples. Previously collected samples are often less than this mass, which means that multiple samples within an experimental treatment must be pooled to reach the minimum requirement. However, samples should be pooled as little as possible to maximize biological replicates.
For example, samples within Treatment A may have these masses: 8g, 7g, 5g, and 10g. Another Treatment, B, has samples with masses of 20g, 21g, 24g, and 29g.
If the minimum mass required for the analysis is 15g, then each sample in Treatment B can be analyzed without pooling. However, in Treatment A, samples would need to be pooled to reach this minimum.
It would be best to combine the 5g and 10g sample and the 8g and 7g sample, because this maximizes the number of possible pooled samples by minimizing potential total masses (i.e., if I combined the 5g and 8g and also the 10g and 7g, I would only have one possible pooled sample that meets the minimum)
Data and R
The data is structured as this example follows:
sample_id = c(1:24)
treatments = c(rep("A",8),rep("B",8),rep("C",8))
mass = round(c(runif(8,4,10),runif(8,5,13),runif(8,15,18)),1)
df = data.frame(cbind(sample_id,treatments,mass))
df$mass = as.numeric(df$mass)
df$sample_id = as.numeric(df$sample_id)
> df
sample_id treatments mass
1 1 A 8.6
2 2 A 8.9
3 3 A 7.5
4 4 A 4.5
5 5 A 7.9
6 6 A 4.5
7 7 A 7.7
8 8 A 6.6
9 9 B 5.0
10 10 B 12.0
11 11 B 7.4
12 12 B 8.4
13 13 B 12.2
14 14 B 10.0
15 15 B 6.5
16 16 B 12.1
17 17 C 15.6
18 18 C 16.5
19 19 C 16.8
20 20 C 17.5
21 21 C 15.6
22 22 C 17.6
23 23 C 18.0
24 24 C 15.8
So far my strategy has been:
# Step 1: separate out all samples that do not need to be pooled, for ease IRL
bigenough = df %>%
filter(mass >= 15)
#Keep df with all the samples that will need to be pooled
poolneeded = df %>%
filter(!(sample_id %in% bigenough$sample_id))
I am at a loss of how to best pool the samples algorithmically however. If anyone has any suggestions that would be helpful. I usually use tidyverse if that helps...
Here is a first attempt. It is made up of a function which takes a split data.frame (split by treatment) of the data to be pooled.
In this function a new DF is created which contains all pairwise possibilities of the sample_id. This df2 is then 2 times left_join-ed with data and the sum of the two samples is calculated, filtered for being >= 15 and ordered.
This function is then called by map after group_split. The result is all the possible allowed sample combinations.
library(tidyverse)
fff <- function(data) {
nn <- nrow(data)
mm <- combn(seq(data$sample_id[1], data$sample_id[nn]), 2) |> t()
df2 <- data.frame(mm) |> setNames(c("sample_id", "sample_id_2"))
ddf <- df2 |>
left_join(data) |> # nolint: object_usage_linter.
left_join(data, by = c("sample_id_2" = "sample_id", "treatments")) |>
mutate(sum = mass.x + mass.y) |> # nolint: object_usage_linter.
filter(sum >= 15) |>
arrange(sample_id, sum) # nolint: object_usage_linter.
return(ddf)
}
poolneeded |>
group_split(treatments) |>
map(fff)
#> Joining, by = "sample_id"
#> Joining, by = "sample_id"
#> [[1]]
#> sample_id sample_id_2 treatments mass.x mass.y sum
#> 1 1 2 A 9.4 6.2 15.6
#> 2 1 7 A 9.4 6.5 15.9
#> 3 1 4 A 9.4 6.8 16.2
#> 4 1 3 A 9.4 7.6 17.0
#> 5 1 8 A 9.4 8.9 18.3
#> 6 2 8 A 6.2 8.9 15.1
#> 7 3 8 A 7.6 8.9 16.5
#> 8 4 8 A 6.8 8.9 15.7
#> 9 7 8 A 6.5 8.9 15.4
#>
#> [[2]]
#> sample_id sample_id_2 treatments mass.x mass.y sum
#> 1 9 10 B 10.9 7.0 17.9
#> 2 9 14 B 10.9 7.2 18.1
#> 3 9 11 B 10.9 7.9 18.8
#> 4 9 16 B 10.9 8.5 19.4
#> 5 9 13 B 10.9 11.2 22.1
#> 6 9 12 B 10.9 11.7 22.6
#> 7 9 15 B 10.9 11.7 22.6
#> 8 10 16 B 7.0 8.5 15.5
#> 9 10 13 B 7.0 11.2 18.2
#> 10 10 12 B 7.0 11.7 18.7
#> 11 10 15 B 7.0 11.7 18.7
#> 12 11 14 B 7.9 7.2 15.1
#> 13 11 16 B 7.9 8.5 16.4
#> 14 11 13 B 7.9 11.2 19.1
#> 15 11 12 B 7.9 11.7 19.6
#> 16 11 15 B 7.9 11.7 19.6
#> 17 12 14 B 11.7 7.2 18.9
#> 18 12 16 B 11.7 8.5 20.2
#> 19 12 13 B 11.7 11.2 22.9
#> 20 12 15 B 11.7 11.7 23.4
#> 21 13 14 B 11.2 7.2 18.4
#> 22 13 16 B 11.2 8.5 19.7
#> 23 13 15 B 11.2 11.7 22.9
#> 24 14 16 B 7.2 8.5 15.7
#> 25 14 15 B 7.2 11.7 18.9
#> 26 15 16 B 11.7 8.5 20.2
Another way
This makes use of the same function fff as above but it needs to be called with a subset of the poolneeeded - in this case below it is a subset of treatments == "B" .
You see then a DF of all possible allowed combinations for pooling and can choose a first pair for pooling. Then the remaining choices for a second pooling are also shown.
sel2 <- function(data) {
ddf <- fff(data)
cat(paste("\n", "These are your possibilities for the FIRST pooling", "\n"))
print(ddf)
ask <- askYesNo("Do You want to make first choice?")
if (ask) {
s_1 <- readline(prompt = "Enter sample 1: ")
s_2 <- readline(prompt = "Enter sample 2: ")
ddf2 <- ddf |> filter(
sample_id != s_1 & sample_id != s_2 &
sample_id_2 != s_1 & sample_id_2 != s_2 # nolint: object_usage_linter.
)
cat(paste0("\n", "These are your possibilities for the SECOND pooling", "\n"))
print(ddf2)
} else {
return()
}
}
poolneeded_b <- poolneeded |> filter(treatments == "B")
sel2(poolneeded_b)
#> r$> sel2(poolneeded_b)
#> Joining, by = "sample_id"
#>
#> These are your possibilities for the FIRST pooling
#> sample_id sample_id_2 treatments mass.x mass.y sum
#> 1 9 14 B 7.6 8.1 15.7
#> 2 9 12 B 7.6 8.7 16.3
#> 3 9 15 B 7.6 9.6 17.2
#> 4 9 10 B 7.6 10.3 17.9
#> 5 9 16 B 7.6 10.9 18.5
#> 6 9 13 B 7.6 12.3 19.9
#> 7 10 11 B 10.3 5.6 15.9
#> 8 10 14 B 10.3 8.1 18.4
#> 9 10 12 B 10.3 8.7 19.0
#> 10 10 15 B 10.3 9.6 19.9
#> 11 10 16 B 10.3 10.9 21.2
#> 12 10 13 B 10.3 12.3 22.6
#> 13 11 15 B 5.6 9.6 15.2
#> 14 11 16 B 5.6 10.9 16.5
#> 15 11 13 B 5.6 12.3 17.9
#> 16 12 14 B 8.7 8.1 16.8
#> 17 12 15 B 8.7 9.6 18.3
#> 18 12 16 B 8.7 10.9 19.6
#> 19 12 13 B 8.7 12.3 21.0
#> 20 13 14 B 12.3 8.1 20.4
#> 21 13 15 B 12.3 9.6 21.9
#> 22 13 16 B 12.3 10.9 23.2
#> 23 14 15 B 8.1 9.6 17.7
#> 24 14 16 B 8.1 10.9 19.0
#> 25 15 16 B 9.6 10.9 20.5
#>
#> Do You want to make first choice? (Yes/no/abbrechen) y
#> Enter sample 1: 9
#> Enter sample 2: 14
#>
#> These are your possibilities for the SECOND pooling
#> sample_id sample_id_2 treatments mass.x mass.y sum
#> 1 10 11 B 10.3 5.6 15.9
#> 2 10 12 B 10.3 8.7 19.0
#> 3 10 15 B 10.3 9.6 19.9
#> 4 10 16 B 10.3 10.9 21.2
#> 5 10 13 B 10.3 12.3 22.6
#> 6 11 15 B 5.6 9.6 15.2
#> 7 11 16 B 5.6 10.9 16.5
#> 8 11 13 B 5.6 12.3 17.9
#> 9 12 15 B 8.7 9.6 18.3
#> 10 12 16 B 8.7 10.9 19.6
#> 11 12 13 B 8.7 12.3 21.0
#> 12 13 15 B 12.3 9.6 21.9
#> 13 13 16 B 12.3 10.9 23.2
#> 14 15 16 B 9.6 10.9 20.5
I'm using qtgrace for MacOS and when I plotted two data in qtgrace I got something like this:
Overlapping data sets
However, I would like to plot something like this:
Non-overlapping data sets
My data 1:
0 14
0.1 6
0.2 14
0.3 14
0.4 14
0.5 14
0.6 14
0.7 14
0.8 6
0.9 6
1 6
1.1 6
1.2 6
1.3 6
1.4 6
1.5 6
1.6 6
1.7 6
1.8 6
1.9 6
2 6
2.1 6
2.2 6
2.3 6
2.4 6
2.5 6
2.6 6
2.7 6
2.8 6
2.9 6
3 6
3.1 6
3.2 6
3.3 6
3.4 6
3.5 6
3.6 6
3.7 6
3.8 6
3.9 6
4 6
4.1 6
4.2 6
4.3 6
4.4 6
4.5 6
4.6 6
4.7 6
4.8 6
4.9 6
5 6
5.1 6
5.2 6
5.3 6
5.4 6
5.5 6
5.6 6
5.7 6
5.8 6
5.9 6
6 6
6.1 6
6.2 6
6.3 6
6.4 6
6.5 6
6.6 6
6.7 6
6.8 6
6.9 6
7 6
7.1 6
7.2 6
7.3 2
7.4 6
7.5 2
7.6 2
7.7 2
7.8 2
7.9 6
8 2
8.1 6
8.2 2
8.3 2
8.4 6
8.5 6
8.6 6
8.7 2
8.8 6
8.9 19
9 19
9.1 6
9.2 6
9.3 6
9.4 2
9.5 2
9.6 2
9.7 2
9.8 2
9.9 2
10 2
10.1 2
10.2 2
10.3 2
10.4 2
10.5 2
10.6 2
10.7 2
10.8 2
10.9 2
11 2
11.1 2
11.2 2
11.3 2
11.4 2
11.5 2
11.6 2
11.7 2
11.8 2
11.9 2
12 2
12.1 2
12.2 2
12.3 2
12.4 2
12.5 2
12.6 2
12.7 2
12.8 2
12.9 2
13 2
13.1 2
13.2 2
13.3 2
13.4 2
13.5 2
13.6 2
13.7 2
13.8 2
13.9 2
14 2
14.1 2
14.2 2
14.3 2
14.4 2
14.5 2
14.6 2
14.7 2
14.8 2
14.9 2
15 2
15.1 2
15.2 2
15.3 2
15.4 2
15.5 2
15.6 2
15.7 2
15.8 2
15.9 2
16 2
16.1 2
16.2 2
16.3 2
16.4 2
16.5 2
16.6 2
16.7 2
16.8 2
16.9 2
17 2
17.1 2
17.2 2
17.3 2
17.4 2
17.5 2
17.6 2
17.7 2
17.8 2
17.9 2
18 2
18.1 2
18.2 2
18.3 2
18.4 2
18.5 2
18.6 2
18.7 2
18.8 2
18.9 2
19 2
19.1 2
19.2 2
19.3 2
19.4 2
19.5 2
19.6 2
19.7 2
19.8 2
19.9 2
20 2
20.1 2
20.2 2
20.3 2
20.4 2
20.5 2
20.6 2
20.7 2
20.8 2
20.9 2
21 2
21.1 2
21.2 2
21.3 2
21.4 2
21.5 2
21.6 2
21.7 2
21.8 7
21.9 2
22 2
22.1 2
22.2 2
22.3 7
22.4 7
22.5 7
22.6 7
22.7 7
22.8 2
22.9 2
23 7
23.1 7
23.2 7
23.3 7
23.4 7
23.5 2
23.6 2
23.7 2
23.8 2
23.9 2
24 2
24.1 2
24.2 2
24.3 2
24.4 2
24.5 2
24.6 2
24.7 2
24.8 2
24.9 2
25 2
. .
. .
. .
Data 2:
0 4
0.1 4
0.2 4
0.3 4
0.4 4
0.5 4
0.6 4
0.7 4
0.8 4
0.9 4
1 2
1.1 4
1.2 4
1.3 4
1.4 4
1.5 4
1.6 4
1.7 4
1.8 4
1.9 4
2 4
2.1 4
2.2 4
2.3 4
2.4 4
2.5 4
2.6 4
2.7 4
2.8 4
2.9 4
3 4
3.1 4
3.2 4
3.3 4
3.4 4
3.5 4
3.6 4
3.7 4
3.8 4
3.9 4
4 4
4.1 4
4.2 4
4.3 4
4.4 4
4.5 4
4.6 4
4.7 4
4.8 4
4.9 4
5 4
5.1 4
5.2 4
5.3 4
5.4 4
5.5 4
5.6 4
5.7 4
5.8 4
5.9 4
6 4
6.1 4
6.2 4
6.3 4
6.4 4
6.5 4
6.6 4
6.7 4
6.8 4
6.9 4
7 4
7.1 4
7.2 4
7.3 4
7.4 4
7.5 4
7.6 4
7.7 4
7.8 4
7.9 4
8 4
8.1 4
8.2 4
8.3 4
8.4 2
8.5 4
8.6 4
8.7 4
8.8 4
8.9 4
9 4
9.1 4
9.2 4
9.3 4
9.4 4
9.5 4
9.6 4
9.7 4
9.8 4
9.9 4
10 4
10.1 4
10.2 4
10.3 4
10.4 4
10.5 2
10.6 2
10.7 4
10.8 2
10.9 2
11 2
11.1 2
11.2 4
11.3 4
11.4 2
11.5 2
11.6 2
11.7 2
11.8 2
11.9 2
12 2
12.1 2
12.2 2
12.3 2
12.4 4
12.5 4
12.6 2
12.7 2
12.8 4
12.9 2
13 2
13.1 4
13.2 4
13.3 4
13.4 4
13.5 10
13.6 2
13.7 2
13.8 2
13.9 2
14 2
14.1 2
14.2 2
14.3 10
14.4 2
14.5 2
14.6 4
14.7 2
14.8 2
14.9 4
15 2
15.1 10
15.2 2
15.3 2
15.4 2
15.5 2
15.6 2
15.7 2
15.8 2
15.9 2
16 2
16.1 2
16.2 2
16.3 2
16.4 2
16.5 2
16.6 2
16.7 2
16.8 2
16.9 2
17 2
17.1 2
17.2 2
17.3 2
17.4 2
17.5 2
17.6 2
17.7 2
17.8 2
17.9 2
18 2
18.1 2
18.2 2
18.3 2
18.4 2
18.5 2
18.6 2
18.7 2
18.8 2
18.9 2
19 2
19.1 2
19.2 2
19.3 2
19.4 2
19.5 2
19.6 2
19.7 2
19.8 2
19.9 2
20 2
20.1 2
20.2 2
20.3 2
20.4 2
20.5 2
20.6 2
20.7 2
20.8 2
20.9 2
21 2
21.1 2
21.2 2
21.3 2
21.4 2
21.5 2
21.6 2
21.7 2
21.8 2
21.9 2
22 2
22.1 2
22.2 2
22.3 2
22.4 2
22.5 2
22.6 2
22.7 2
22.8 2
22.9 2
23 2
23.1 2
23.2 2
23.3 2
23.4 2
23.5 2
23.6 2
23.7 2
23.8 2
23.9 2
24 2
24.1 2
24.2 2
24.3 2
24.4 2
24.5 2
24.6 2
24.7 2
24.8 2
24.9 2
25 2
. .
. .
. .
The data are in two separate xvg file from GROMACS cluster analysis. I wanna plot five different sets in a manner which I can see all data without superposing.
Thank you!
I think the best approach would be to write a script that takes the original files and spits out new files with shifted y values. However, since you have asked for a qt/xmgrace solution, here is how you do it:
Load up all the datasets into qtgrace
Open the "Data -> Transformations -> Evaluate expression..." dialog
Select in the left and right columns a dataset and in the textbox below enter the formula y = y + 0.1. Click "apply". This will shift the dataset up by 0.1
Select the next dataset in the same way and use the formula y = y + 0.2. Click apply
Rinse and repeat for all the datasets (changing the shift accordingly)
When I check the type, it said "integer", but has decimal point. If I change it to numeric, it become integer(no decimal point).
Because I want to do histogram, x must be numeric, but if change to numeric, all data wrong.
> typeof(data$fare_amount)
[1] "integer"
> data$fare_amount
[1] 5.5 6.5 8.0 13.5 5.5 9.5 7.5 8.0 16.0 8.0 5.5 7.0 8.0 5.0 9.5 23.0 5.0 6.0 17.5 12.0 8.5 13.0
[23] 6.5 4.5 52.0 14.5 7.5 4.5 9.0 10.0 15.0 11.5 6.0 12.5 7.5 8.0 6.5 7.5 31.5 10.0 10.0 10.0 4.0 8.5
[45] 24.0 8.5 5.5 14.0 11.0 4.5 9.0 7.5 22.0 8.5 24.0 36.5 15.0 10.5 9.5 17.0 4.5 6.0 6.5 11.5 16.0 6.5
[67] 7.0 20.0 13.5 30.0 8.0 11.0 6.5 11.5 6.5 37.0 5.5 12.5 8.5 58.5 13.5 8.5 9.0 6.0 6.5 9.0 38.0 4.5
[89] 10.0 9.0 44.5 11.0 12.0 4.5 14.5 8.5 32.0 9.5 4.5 6.0 6.5 6.0 31.5 52.0 10.5 12.0 5.5 24.5 7.0 5.5
[111] 16.5 5.0 5.5 6.5 3.5 11.5 13.0 6.0 14.0 3.5
42 Levels: 13.5 16.0 5.5 6.5 7.5 8.0 9.5 12.0 17.5 23.0 5.0 6.0 7.0 10.0 13.0 14.5 4.5 52.0 8.5 9.0 11.5 12.5 ... 3.5
> temp <- as.numeric(data$fare_amount)
> temp
[1] 3 4 6 1 3 7 5 6 2 6 3 13 6 11 7 10 11 12 9 8 19 15 4 17 18 16 5 17 20 14 23 21 12 22 5 6 4 5
[39] 24 14 14 14 28 19 27 19 3 26 25 17 20 5 31 19 27 32 23 29 7 30 17 12 4 21 2 4 13 33 1 34 6 25 4 21 4 35
[77] 3 22 19 36 1 19 20 12 4 20 37 17 14 20 39 25 8 17 16 19 38 7 17 12 4 12 24 18 29 8 3 40 13 3 41 11 3 4
[115] 42 21 15 12 26 42
I was expecting points but got this when I did
plot(data$v3,data$v2)
my data
V2 V3
2 -2.0 2.7
3 0.5 3.9
4 1.3 4.5
5 5.7 6.0
6 10.4 8.7
7 3.4 2.7
8 7.6 3.2
9 4.1 5.6
10 5.0 9.2
11 8.5 11.7
12 12.3 6.8
13 16.1 13.0
14 13.2 11.9
15 8.8 8.6
16 7.9 6.1
17 1.1 4.9
18 3.0 1.0
19 4.5 7.2
20 2.7 2.7
21 7.6 7.6
I tried searching but from my understanding the function is supposed to give points, not bars. How do I fix this?
Is there a way that one can set the number of digits of a full data frame to 2? Case in point, how would you set the number of digits to 2 for the following data using R?
Distance Age Height Coning
1 21.4 18 3.3 Yes
2 13.9 17 3.4 Yes
3 23.9 16 2.9 Yes
4 8.7 18 3.6 No
5 241.8 6 0.7 No
6 44.5 17 1.3 Yes
7 30.0 15 2.5 Yes
8 32.3 16 1.8 Yes
9 31.4 17 5.0 No
10 32.8 13 1.6 No
11 53.3 12 2.0 No
12 54.3 6 0.9 No
13 96.3 11 2.6 No
14 133.6 4 0.6 No
15 32.1 15 2.3 No
16 57.9 12 2.4 Yes
17 30.8 17 1.8 No
18 59.9 7 0.8 No
19 42.7 15 2.0 Yes
20 20.6 18 1.7 Yes
21 62.0 8 1.3 No
22 53.1 7 1.6 No
23 28.9 16 2.2 Yes
24 177.4 5 1.1 No
25 24.8 14 1.5 Yes
26 75.3 14 2.3 Yes
27 51.6 7 1.4 No
28 36.1 9 1.1 No
29 116.1 6 1.1 No
30 28.1 16 2.5 Yes
31 8.7 19 2.2 Yes
32 105.1 6 0.8 No
33 46.0 15 3.0 Yes
34 102.6 7 1.2 No
35 15.8 15 2.2 No
36 60.0 7 1.3 No
37 96.4 13 2.6 No
38 24.2 14 1.7 No
39 14.5 15 2.4 No
40 36.6 14 1.5 No
41 65.7 5 0.6 No
42 116.3 7 1.6 No
43 113.6 8 1.0 No
44 16.7 15 4.3 Yes
45 66.0 7 1.0 No
46 60.7 7 1.0 No
47 90.6 7 0.7 No
48 91.3 7 1.3 No
49 14.4 18 3.1 Yes
50 72.8 14 3.0 Yes
It sounds like you're just looking for format, which has a data.frame method.
A small example:
mydf <- mydf2 <- data.frame(
Distance = c(21.4, 13.9, 23.9, 8.7, 241.8, 44.5),
Age = c(18, 17, 16, 18, 6, 17),
Height = c(3.3, 3.4, 2.9, 3.6, 0.7, 1.3),
Coning = c("Y", "Y", "Y", "N", "N", "Y"))
format(mydf, nsmall = 2)
# Distance Age Height Coning
# 1 21.40 18.00 3.30 Y
# 2 13.90 17.00 3.40 Y
# 3 23.90 16.00 2.90 Y
# 4 8.70 18.00 3.60 N
# 5 241.80 6.00 0.70 N
# 6 44.50 17.00 1.30 Y
As you should expect, if the data are integers, they won't be printed as decimals.
mydf2$Age <- as.integer(mydf2$Age)
format(mydf2, nsmall = 2)
# Distance Age Height Coning
# 1 21.40 18 3.30 Y
# 2 13.90 17 3.40 Y
# 3 23.90 16 2.90 Y
# 4 8.70 18 3.60 N
# 5 241.80 6 0.70 N
# 6 44.50 17 1.30 Y
An alternative to format is to globally set the option for the number of digits to be displayed:
options(digits=2)
Will mean that from that point forward all numerics will be printed up to 2 decimal places (The default is 7).