Calculating percentages/conditionally summing with stat_summary_hex
I have a set of lat/long values with a third variable that can be either -1 or 1. (Sample data: https://pastebin.com/WPq90ka1)
I am mapping this data into hex bins using stat_summary_hex:
single <- read.csv(file="mysamplefile.csv",head=TRUE)
pdxMap <- get_stamenmap(bbox = c(left = -122.844435, bottom = 45.42, right =
-122.43, top = 45.665316), zoom = 11, maptype = "toner")
ggmap(pdxMap) +
coord_equal() + theme(aspect.ratio = 1) +
stat_summary_hex(data=single, aes(x=LONGITUDE,y=LATITUDE,z = Value),
fun = function(z) mean(z),
alpha = .9, bins=50) +
scale_fill_gradientn(colors=c('#4393c3', '#ffffff', '#d6604d'))
However, instead of doing mean(z), I'd prefer to calculate the percentage of values that are 1, ie: (sum of values that=1)/(sum of all abs(values))
The denominator isn't an issue, but I can't figure out how to manage the numerator. Any guidance would be much appreciated!
r ggplot2 ggmap
asked Nov 22 '18 at 2:00
user3754738user3754738
62
add a comment |
I have a set of lat/long values with a third variable that can be either -1 or 1. (Sample data: https://pastebin.com/WPq90ka1)
I am mapping this data into hex bins using stat_summary_hex:
single <- read.csv(file="mysamplefile.csv",head=TRUE)
pdxMap <- get_stamenmap(bbox = c(left = -122.844435, bottom = 45.42, right =
-122.43, top = 45.665316), zoom = 11, maptype = "toner")
ggmap(pdxMap) +
coord_equal() + theme(aspect.ratio = 1) +
stat_summary_hex(data=single, aes(x=LONGITUDE,y=LATITUDE,z = Value),
fun = function(z) mean(z),
alpha = .9, bins=50) +
scale_fill_gradientn(colors=c('#4393c3', '#ffffff', '#d6604d'))
However, instead of doing mean(z), I'd prefer to calculate the percentage of values that are 1, ie: (sum of values that=1)/(sum of all abs(values))
The denominator isn't an issue, but I can't figure out how to manage the numerator. Any guidance would be much appreciated!
r ggplot2 ggmap
asked Nov 22 '18 at 2:00
user3754738user3754738
62
Found a solution using ifelse: fun = function(z) sum(ifelse(z>0,z,0))/sum(abs(z)))
– user3754738
Nov 22 '18 at 2:32
1
Nothing wrong with your solution, but a more readable solution takes advantage of the fact thatas.numeric(T) == 1andas.numeric(F) == 0. Given this, you could usesum(z==1) / length(z)
– Geoffrey Poole
Nov 22 '18 at 5:16
add a comment |
I have a set of lat/long values with a third variable that can be either -1 or 1. (Sample data: https://pastebin.com/WPq90ka1)
I am mapping this data into hex bins using stat_summary_hex:
single <- read.csv(file="mysamplefile.csv",head=TRUE)
pdxMap <- get_stamenmap(bbox = c(left = -122.844435, bottom = 45.42, right =
-122.43, top = 45.665316), zoom = 11, maptype = "toner")
ggmap(pdxMap) +
coord_equal() + theme(aspect.ratio = 1) +
stat_summary_hex(data=single, aes(x=LONGITUDE,y=LATITUDE,z = Value),
fun = function(z) mean(z),
alpha = .9, bins=50) +
scale_fill_gradientn(colors=c('#4393c3', '#ffffff', '#d6604d'))
However, instead of doing mean(z), I'd prefer to calculate the percentage of values that are 1, ie: (sum of values that=1)/(sum of all abs(values))
The denominator isn't an issue, but I can't figure out how to manage the numerator. Any guidance would be much appreciated!
r ggplot2 ggmap
asked Nov 22 '18 at 2:00
user3754738user3754738
62
I have a set of lat/long values with a third variable that can be either -1 or 1. (Sample data: https://pastebin.com/WPq90ka1)
I am mapping this data into hex bins using stat_summary_hex:
single <- read.csv(file="mysamplefile.csv",head=TRUE)
pdxMap <- get_stamenmap(bbox = c(left = -122.844435, bottom = 45.42, right =
-122.43, top = 45.665316), zoom = 11, maptype = "toner")
ggmap(pdxMap) +
coord_equal() + theme(aspect.ratio = 1) +
stat_summary_hex(data=single, aes(x=LONGITUDE,y=LATITUDE,z = Value),
fun = function(z) mean(z),
alpha = .9, bins=50) +
scale_fill_gradientn(colors=c('#4393c3', '#ffffff', '#d6604d'))
However, instead of doing mean(z), I'd prefer to calculate the percentage of values that are 1, ie: (sum of values that=1)/(sum of all abs(values))
The denominator isn't an issue, but I can't figure out how to manage the numerator. Any guidance would be much appreciated!
r ggplot2 ggmap
r ggplot2 ggmap
asked Nov 22 '18 at 2:00
user3754738user3754738
62
asked Nov 22 '18 at 2:00
user3754738user3754738
62
asked Nov 22 '18 at 2:00
user3754738user3754738
62
asked Nov 22 '18 at 2:00
user3754738user3754738
62
asked Nov 22 '18 at 2:00
user3754738user3754738
62
62
Found a solution using ifelse: fun = function(z) sum(ifelse(z>0,z,0))/sum(abs(z)))
– user3754738
Nov 22 '18 at 2:32
1
Nothing wrong with your solution, but a more readable solution takes advantage of the fact thatas.numeric(T) == 1andas.numeric(F) == 0. Given this, you could usesum(z==1) / length(z)
– Geoffrey Poole
Nov 22 '18 at 5:16
add a comment |
Found a solution using ifelse: fun = function(z) sum(ifelse(z>0,z,0))/sum(abs(z)))
– user3754738
Nov 22 '18 at 2:32
1
Nothing wrong with your solution, but a more readable solution takes advantage of the fact thatas.numeric(T) == 1andas.numeric(F) == 0. Given this, you could usesum(z==1) / length(z)
– Geoffrey Poole
Nov 22 '18 at 5:16
Found a solution using ifelse: fun = function(z) sum(ifelse(z>0,z,0))/sum(abs(z)))
– user3754738
Nov 22 '18 at 2:32
Found a solution using ifelse: fun = function(z) sum(ifelse(z>0,z,0))/sum(abs(z)))
– user3754738
Nov 22 '18 at 2:32
1
1
Nothing wrong with your solution, but a more readable solution takes advantage of the fact that
as.numeric(T) == 1 and as.numeric(F) == 0. Given this, you could use sum(z==1) / length(z)– Geoffrey Poole
Nov 22 '18 at 5:16
Nothing wrong with your solution, but a more readable solution takes advantage of the fact that
as.numeric(T) == 1 and as.numeric(F) == 0. Given this, you could use sum(z==1) / length(z)– Geoffrey Poole
Nov 22 '18 at 5:16
add a comment |
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Found a solution using ifelse: fun = function(z) sum(ifelse(z>0,z,0))/sum(abs(z)))
– user3754738
Nov 22 '18 at 2:32
1
Nothing wrong with your solution, but a more readable solution takes advantage of the fact that
as.numeric(T) == 1andas.numeric(F) == 0. Given this, you could usesum(z==1) / length(z)– Geoffrey Poole
Nov 22 '18 at 5:16