I'm having difficult time to crate a two-way frequency distribution table for a tree diameter-height data. I have a dataset like below
dbh(cm) tht(m)
3 53.35
19 13.37
27 16.53
22 17.8
9 8.33
10 8.76
24 15.62
44 30.3
17 14.91
10 8.93
I need to create a frequency distribution table for this data with classes for both columns. My class bounds are like,
for diameter:
8 - 11.9
12 - 15.9
16 - 19.9 and so.
for height:
3 - 4.9
5 - 6.9
7 - 8.9 and so.
So I have thousands of rows of data and summing up each frequency in itself is a total pain. I wrote the following lines (dbh represents diameter and tht is height);
> data <- read.csv('data.csv')
> diameter <- data$dbh
> range(diameter)
[1] 6.0 60.5
> breaks <- seq(6, 61, by=4)
> diameter.cut <- cut(diameter, breaks, right = FALSE)
> diameter.frq <- table(diameter.cut)
> cbind(diameter.frq)
diameter.frq
[6,10) 35
[10,14) 77
[14,18) 59
[18,22) 25
[22,26) 25
[26,30) 51
[30,34) 38
[34,38) 28
[38,42) 21
[42,46) 22
[46,50) 14
[50,54) 2
[54,58) 6
>
I can do the same thing to the tree height (tht). But the problem is how do I create a 'cross frequence table'
For example: there are 35 trees in [6,10) diameter class. But I need to distribute these 35 trees into each height class. Let's say 12 of these trees belong to [3,5) height class, 8 of them in [5,7)...etc.
P.S: I'm pretty new in R. So my problem might seem pretty dummy but I really digged the internet before I post here. I'm sorry for that.
Did you mean this? After tweaking your code I got...
library(dplyr)
df %>%
mutate(diameter.cm = cut(diameter.cm, seq(range(diameter.cm)[1], range(diameter.cm)[2]+4, by=4), right = F),
height.m = cut(height.m, seq(range(height.m)[1], range(height.m)[2]+2, by=2), right = F)) %>%
group_by(diameter.cm, height.m) %>%
tally()
Output is:
diameter.cm height.m n
1 [3,7) [52.3,54.3) 1
2 [7,11) [8.33,10.3) 3
3 [15,19) [14.3,16.3) 1
4 [19,23) [12.3,14.3) 1
5 [19,23) [16.3,18.3) 1
6 [23,27) [14.3,16.3) 1
7 [27,31) [16.3,18.3) 1
8 [43,47) [28.3,30.3) 1
Sample data:
df <- structure(list(diameter.cm = c(3L, 19L, 27L, 22L, 9L, 10L, 24L,
44L, 17L, 10L), height.m = c(53.35, 13.37, 16.53, 17.8, 8.33,
8.76, 15.62, 30.3, 14.91, 8.93)), class = "data.frame", row.names = c(NA,
-10L))