Re: [R] Help with plotting and date-times for climate data
Às 21:50 de 12/09/2023, Kevin Zembower via R-help escreveu:
Hello,
I'm trying to calculate the mean temperature max from a file of climate
date, and plot it over a range of days in the year. I've downloaded the
data, and cleaned it up the way I think it should be. However, when I
plot it, the geom_smooth line doesn't show up. I think that's because
my x axis is characters or factors. Here's what I have so far:
library(tidyverse)
data <- read_csv("Ely_MN_Weather.csv")
start_day = yday(as_date("2023-09-22"))
end_day = yday(as_date("2023-10-15"))
d <- as_tibble(data) %>%
select(DATE,TMAX,TMIN) %>%
mutate(DATE = as_date(DATE),
yday = yday(DATE),
md = sprintf("%02d-%02d", month(DATE), mday(DATE))
) %>%
filter(yday >= start_day & yday <= end_day) %>%
mutate(md = as.factor(md))
d_sum <- d %>%
group_by(md) %>%
summarize(tmax_mean = mean(TMAX, na.rm=TRUE))
## Here's the filtered data:
dput(d_sum)
structure(list(md = structure(1:25, levels = c("09-21", "09-22",
"09-23", "09-24", "09-25", "09-26", "09-27", "09-28", "09-29",
"09-30", "10-01", "10-02", "10-03", "10-04", "10-05", "10-06",
"10-07", "10-08", "10-09", "10-10", "10-11", "10-12", "10-13",
"10-14", "10-15"), class = "factor"), tmax_mean = c(65,
62.2,
61.3, 63.9, 64.3, 60.1, 62.3, 60.5, 61.9,
61.2, 63.7, 59.5, 59.6, 61.6,
59.4, 58.8, 55.9, 58.125,
58, 55.7, 57, 55.4, 49.8,
48.75, 43.7)), class = c("tbl_df", "tbl", "data.frame"
), row.names = c(NA, -25L))
ggplot(data = d_sum, aes(x = md)) +
geom_point(aes(y = tmax_mean, color = "blue")) +
geom_smooth(aes(y = tmax_mean, color = "blue"))
=
My questions are:
1. Why isn't my geom_smooth plotting? How can I fix it?
2. I don't think I'm handling the month and day combination correctly.
Is there a way to encode month and day (but not year) as a date?
3. (Minor point) Why does my graph of tmax_mean come out red when I
specify "blue"?
Thanks for any advice or guidance you can offer. I really appreciate
the expertise of this group.
-Kevin
__
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Hello,
The problem is that the dates are factors, not real dates. And
geom_smooth is not interpolating along a discrete axis (the x axis).
Paste a fake year with md, coerce to date and plot.
I have simplified the aes() calls and added a date scale in order to
make the x axis more readable.
Without the formula and method arguments, geom_smooth will print a
message, they are now made explicit.
suppressPackageStartupMessages({
library(dplyr)
library(ggplot2)
})
d_sum %>%
mutate(md = paste("2023", md, sep = "-"),
md = as.Date(md)) %>%
ggplot(aes(x = md, y = tmax_mean)) +
geom_point(color = "blue") +
geom_smooth(
formula = y ~ x,
method = loess,
color = "blue"
) +
scale_x_date(date_breaks = "7 days", date_labels = "%m-%d")
Hope this helps,
Rui Barradas
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Re: [R] graph in R with grouping letters from the turkey test with agricolae package
Às 16:24 de 12/09/2023, Loop Vinyl escreveu: I would like to produce the attached graph (graph1) with the R package agricolae, could someone give me an example with the attached data (data)? I expect an adapted graph (graph2) with the data (data) Best regards __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Hello, There are no attached graphs, only data. Can you post the code have you tried? Hope this helps, Rui Barradas __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Help with plotting and date-times for climate data
Hello,
I'm trying to calculate the mean temperature max from a file of climate
date, and plot it over a range of days in the year. I've downloaded the
data, and cleaned it up the way I think it should be. However, when I
plot it, the geom_smooth line doesn't show up. I think that's because
my x axis is characters or factors. Here's what I have so far:
library(tidyverse)
data <- read_csv("Ely_MN_Weather.csv")
start_day = yday(as_date("2023-09-22"))
end_day = yday(as_date("2023-10-15"))
d <- as_tibble(data) %>%
select(DATE,TMAX,TMIN) %>%
mutate(DATE = as_date(DATE),
yday = yday(DATE),
md = sprintf("%02d-%02d", month(DATE), mday(DATE))
) %>%
filter(yday >= start_day & yday <= end_day) %>%
mutate(md = as.factor(md))
d_sum <- d %>%
group_by(md) %>%
summarize(tmax_mean = mean(TMAX, na.rm=TRUE))
## Here's the filtered data:
dput(d_sum)
> structure(list(md = structure(1:25, levels = c("09-21", "09-22",
"09-23", "09-24", "09-25", "09-26", "09-27", "09-28", "09-29",
"09-30", "10-01", "10-02", "10-03", "10-04", "10-05", "10-06",
"10-07", "10-08", "10-09", "10-10", "10-11", "10-12", "10-13",
"10-14", "10-15"), class = "factor"), tmax_mean = c(65,
62.2,
61.3, 63.9, 64.3, 60.1, 62.3, 60.5, 61.9,
61.2, 63.7, 59.5, 59.6, 61.6,
59.4, 58.8, 55.9, 58.125,
58, 55.7, 57, 55.4, 49.8,
48.75, 43.7)), class = c("tbl_df", "tbl", "data.frame"
), row.names = c(NA, -25L))
>
ggplot(data = d_sum, aes(x = md)) +
geom_point(aes(y = tmax_mean, color = "blue")) +
geom_smooth(aes(y = tmax_mean, color = "blue"))
=
My questions are:
1. Why isn't my geom_smooth plotting? How can I fix it?
2. I don't think I'm handling the month and day combination correctly.
Is there a way to encode month and day (but not year) as a date?
3. (Minor point) Why does my graph of tmax_mean come out red when I
specify "blue"?
Thanks for any advice or guidance you can offer. I really appreciate
the expertise of this group.
-Kevin
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] graph in R with grouping letters from the turkey test with agricolae package
I would like to produce the attached graph (graph1) with the R package agricolae, could someone give me an example with the attached data (data)? I expect an adapted graph (graph2) with the data (data) Best regards TREAT REP VAR1VAR2VAR3VAR4 t1 1 16,10,805 0,966 1,449 t1 2 14,10,705 0,846 1,269 t1 3 5,8 0,290,348 0,522 t1 4 4,9 0,245 0,294 0,441 t2 1 30,81,541,848 2,772 t2 2 29 1,451,742,61 t2 3 28,81,441,728 2,592 t2 4 16,60,830,996 1,494 t3 1 31,31,565 1,878 2,817 t3 2 29,61,481,776 2,664 t3 3 28,11,405 1,686 2,529 t3 4 19,50,975 1,171,755 t4 1 21,45 1,0725 1,287 1,9305 t4 2 23,595 1,17975 1,4157 2,12355 t4 3 25,9545 1,2977251,55727 2,335905 t4 4 28,549951,4274975 1,7129972,5694955 t5 1 27,31,365 1,638 2,457 t5 2 13,70,685 0,822 1,233 t5 3 8,1 0,405 0,486 0,729 t5 4 0 0 0 0 __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] prop.trend.test
Argh, yes, drats, thanks. There will be a matter of an estimated residual error. So > coef(summary(ht$lmfit))["score","t value"]*sigma(ht$lmfit) [1] -2.867913 matches the signes square root of the Chi-square. Or, likely better (avoid 0 df cases), switch to a Gaussian glm fit and use the z stat > coef(summary(ht$glmfit, dispersion = 1)) Estimate Std. Error z value Pr(>|z|) (Intercept) 1.02187141 0.03199737 31.936102 8.425506e-224 score -0.03341563 0.01165155 -2.867913 4.131897e-03 The Estimate (-0.0334) should still make sense as the LPM estimate of the regression slope. - Peter D. > On 8 Sep 2023, at 12:53 , Rui Barradas wrote: > > Às 10:06 de 08/09/2023, peter dalgaard escreveu: >> Yes, this was written a bit bone-headed (as I am allowed to say...) >> If you look at the code, you will see inside: >> a <- anova(lm(freq ~ score, data = list(freq = x/n, score = >> as.vector(score)), >> weights = w)) >> and the lm() inside should give you the direction via the sign of the >> regression coefficient on "score". >> So, at least for now, you could just doctor a copy of the code for your own >> purposes, as in >> fit <- lm(freq ~ score, data = list(freq = x/n, score = as.vector(score)), >> weights = w) >> a <- anova(fit) >> and arrange to return coef(fit)["score"] at the end. Something like >> structure(... estimate=c(lpm.slope=coef(fit)["score"]) ) >> (I expect that you might also extract the t-statistic from >> coef(summary(fit)) and find that it is the signed square root of the >> Chi-square, but I won't have time to test that just now.) >> -pd >>> On 8 Sep 2023, at 07:22 , Thomas Subia via R-help >>> wrote: >>> >>> Colleagues, >>> >>> Thanks all for the responses. >>> >>> I am monitoring the daily total number of defects per sample unit. >>> I need to know whether this daily defect proportion is trending upward (a >>> bad thing for a manufacturing process). >>> >>> My first thought was to use either a u or a u' control chart for this. >>> As far as I know, u or u' charts are poor to detect drifts. >>> >>> This is why I chose to use prop.trend.test to detect trends in proportions. >>> >>> While prop.trend.test can confirm the existence of a trend, as far as I >>> know, it is left to the user >>> to determine what direction that trend is. >>> >>> One way to illustrate trending is of course to plot the data and use >>> geom_smooth and method lm >>> For the non-statisticians in my group, I've found that using this method >>> along with the p-value of prop.trend.test, makes it easier for the users to >>> determine the existence of trending and its direction. >>> >>> If there are any other ways to do this, please let me know. >>> >>> Thomas Subia >>> >>> >>> >>> >>> >>> >>> >>> >>> >>> >>> >>> >>> On Thursday, September 7, 2023 at 10:31:27 AM PDT, Rui Barradas >>> wrote: >>> >>> >>> >>> >>> >>> Às 14:23 de 07/09/2023, Thomas Subia via R-help escreveu: Colleagues Consider smokers <- c( 83, 90, 129, 70 ) patients <- c( 86, 93, 136, 82 ) prop.trend.test(smokers, patients) Output: Chi-squared Test for Trend inProportions data: smokers out of patients , using scores: 1 2 3 4 X-squared = 8.2249, df = 1, p-value = 0.004132 # trend test for proportions indicates proportions aretrending. How does one identify the direction of trending? # prop.test indicates that the proportions are unequal but doeslittle to indicate trend direction. All the best, Thomas Subia [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. >>> Hello, >>> >>> By visual inspection it seems that there is a decreasing trend. >>> Note that the sample estimates of prop.test and smokers/patients are equal. >>> >>> >>> smokers <- c( 83, 90, 129, 70 ) >>> patients <- c( 86, 93, 136, 82 ) >>> >>> prop.test(smokers, patients)$estimate >>> #>prop 1prop 2prop 3prop 4 >>> #> 0.9651163 0.9677419 0.9485294 0.8536585 >>> >>> smokers/patients >>> >>> #> [1] 0.9651163 0.9677419 0.9485294 0.8536585 >>> >>> plot(smokers/patients, type = "b") >>> >>> >>> >>> Hope this helps, >>> >>> Rui Barradas >>> >>> __ >>> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see >>> https://stat.ethz.ch/mailman/listinfo/r-help >>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >>> and provide commented, minimal, self-contained, reproducible code. > Hello, > >
