1170:Lab2
August 28th, 2012
Goals For This Week
In this second week we will get some more practice plotting functions. Hopefully, you paid attention
to what you were doing as we’re going to add in a few twists.
• Plot data points and curves on the same plot
• Make our lives easier (in the long run) by defining our own functions
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Plotting data
As scientists, we usually want to compare real-world data with a simplified model.
As a quick example consider the mean monthly temperature in Salt Lake City.
Month
Temperature(F)
January
37
February
43
March
53
April
61
May
71
June
82
July
91
August
89
September
78
October
64
November
49
December
38
First let’s plot these data. You can call the variables anything you want; I try to make my names as
descriptive as possible.
> t_data_months<-seq(.5,11.5)
> temp_data_F<-c(37,43,53,61,71,82,91,89,78,64,49,38)
> plot(t_data_months,temp_data_F,xlab=’months’,ylab=’temp(F)’,main=’Temp in SLC’)
The t vector is defined that way because we want to put the data point at the middle of the month.
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Plotting a function on top of data
Next I’m going to come up with a simple function that goes through these points. Let’s assume that is has
the following form.
T = Tavg − Tchangecos(2π(t − t0 )/12)
The temperature range is 37 to 91 degrees so I’m going to guess that Tavg = (91 + 37)/2 = 64 and
Tchange = (91 − 37)/2 = 27. This just leaves t0 unknown.
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Now it’s time to introduce a new and very important command in R. The function command allows us
to define our own functions for later use. We’ll learn how to use this function by example.
> f<-function(t){64-27*cos(2*pi*t/12)}
We can try entering in numbers
> f(3)
or lists of numbers
> f(t_data_months)
Next I’m going to try to put this line on top of the plotted points from before. As I’m plotting a
continuous function I want to create some more t values to make the curve smoother.
> t_model<-seq(0,12,.01)
> lines(t_model,f(t_model))
This doesn’t look quite right, so it would be nice if we could make it a little better. Recall that we left
t0 unknown. We can try varying this parameter to make the curve match the points better. First I’ll guess
that t0 = 1 and then redo the lines command.
> lines(t_model,f(t_model-1))
Now you should see the points along with the two curves plotted on top of each other. Play around a little
with t0 to get the best value. You can clear the picture and start over by re-entering the plot command.
Once you have a good value for t0 , save the picture. The picture should have the original data
points and a single smooth curve going through it that represents your best fit to the data.
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Changing Units
Note in my examples below I’m using t0 = 1, you should use whatever t0 you decided was best.
Now that we have the function defined, let’s try some changes of units. Currently, your function takes
an input in months and gives an output in fahrenheit. Let’s suppose that instead, someone wanted a plot
for the first 12 days of the year. We can reuse the same t values (in fact, we have to), but now their unit is
days. We have to make the conversion from days to months before inputting them to the function.
Now I’m going to add in a factor (1/30) which represents the approximate conversion from months to
days.
> plot(t_model,f(t_model/30-1),xlab=’days’,ylab=’temp(F)’,main=’Temp in SLC’,type=’l’)
Note how multiplying by a number smaller than 1 resulted in a horizontal shrink. Save Your Graph
Now, let’s imagine that we want to display the temperature in celsius.
> plot(t_model,(f(t_model-1)-32)*5/9,xlab=’months’,ylab=’temp(C)’,main=’Temp in SLC’,type=’l’)
Save this plot.
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Assignment for this week
1. A mean monthly temperature in SLC (in F) along with a smooth function that approximates the data.
(Should already be done)
2. Plot your smooth function from the last part for the 1st twelve days of the year. No need to include
the data. (Should already be done)
3. Plot the smooth function for the entire year with months on the horizontal axis and temperature in
celsius on the vertical. (Should already be done)
4. Plot the smooth function for 12 years with years on the horizontal and temperature in Kelvin on the
vertical axis. Note that we convert celsius to Kelvin by adding 273.15.
5. The following table shows the population of Utah. Plot it with the year on the horizontal axis and
population on the vertical axis.
Year
1850
1860
1870
1880
1890
1900
1910
1920
1930
1940
1950
1960
1970
1980
1990
2000
2010
Population
11,380
40,273
86,336
143,963
210,779
276,749
373,351
449,396
507,847
550,310
688,862
890,627
1,059,273
1,461,037
1,722,850
2,223,169
2,763,885
Try to find an exponential function of the form P = A ∗ exp(r ∗ (t − 1850)) that fits this data. Save a
plot that includes the original data and your smooth function.
Hint: Population= g(A, r, t) = Aert . You can encode this into R like this.
> g<-function(A,r,t){A*exp(r*(t-1850))}
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