Say, I have a data.frame() like this
>head(Acquisition)
original_date first_payment_date LTV DTI FICO
1 01/2007 03/2007 56 37 734
2 02/2007 04/2007 80 11 762
3 12/2006 02/2007 80 28 656
4 12/2006 03/2007 70 50 700
I want to discretize the Acquisition$LTV and Acquisition$DTI by the step size 0.05 and Acquisition$FICO by the step size 10.
I have found the answer just use cut function is okay.
dis.LTV=cut(Acquisition$LTV,(max(Acquisition$LTV)-min(Acquisition$LTV))/0.05)
Related
I am attempting to work with a large dataset in R where I need to create a column that compares the value in an existing column to all values that follow it (ex: row 1 needs to compare rows 1-10,000, row 2 needs to compare rows 2-10,000, row 3 needs to compare rows 3-10,000, etc.), but cannot figure out how to write the range.
I currently have a column of raw numeric values and a column of row values generated by:
samples$row = seq.int(nrow(samples))
I have attempted to generate the column with the following command:
samples$processed = min(samples$raw[samples$row:10000])
but get the error "numerical expression has 10000 elements: only the first used" and the generated column only has the value for row 1 repeated for each of the 10,000 rows.
How do I need to write this command so that the lower bound of the range is the row currently being calculated instead of 1?
Any help would be appreciated, as I have minimal programming experience.
If all you need is the min of the specific row and all following rows, then
rev(cummin(rev(samples$val)))
# [1] 24 24 24 24 24 24 24 24 24 24 24 24 165 165 165 165 410 410 410 882
If you have some other function that doesn't have a cumulative variant (and your use of min is just a placeholder), then one of:
mapply(function(a, b) min(samples$val[a:b]), seq.int(nrow(samples)), nrow(samples))
# [1] 24 24 24 24 24 24 24 24 24 24 24 24 165 165 165 165 410 410 410 882
sapply(seq.int(nrow(samples)), function(a) min(samples$val[a:nrow(samples)]))
The only reason to use mapply over sapply is if, for some reason, you want window-like operations instead of always going to the bottom of the frame. (Though if you wanted windows, I'd suggest either the zoo or slider packages.)
Data
set.seed(42)
samples <- data.frame(val = sample(1000, size=20))
samples
# val
# 1 561
# 2 997
# 3 321
# 4 153
# 5 74
# 6 228
# 7 146
# 8 634
# 9 49
# 10 128
# 11 303
# 12 24
# 13 839
# 14 356
# 15 601
# 16 165
# 17 622
# 18 532
# 19 410
# 20 882
I have 20 intervals:
10 intervals from 1 to 250 of size 25:
[1.25] [26.50] [51.75] [76.100] [101.125] [126.150] ... [226.250]
10 intervals from 251 to 1000 of size 75:
[251,325] [326,400] [401,475] [476,550] [551,625] ... [926,1000]
I would like to create a vector composed of the first 5 elements of each interval like:
(1,2,3,5, 26,27,28,29,30, 51,52,53,54,55, 76,77,78,79,80, ....,
251,252,253,254,255, 326,327,328,329,330, ...)
How create this vector using R?
Let's assume you have two interval like :
interval1 <- seq(1.25, 226.250, 25)
interval2 <- seq(251, 1000, 75)
We can create a new interval combining the two and then use mapply to create sequence
new_interval <- c(as.integer(interval1), interval2)
c(mapply(`:`, new_interval, new_interval + 4))
#[1] 1 2 3 4 5 26 27 28 29 30 51 52 53 54 .....
#[89] ..... 779 780 851 852 853 854 855 926 927 928 929 930
I am working with dplyr and sample_n in R and trying to get an even group of rows to work with in my data frame.
So, I have a data set, head of data as follows:
> head(SEH)
Time.Level Demo.Age SEH.Total
92 PRE 12 110
335 PRE 12 80
720 MID 14 85
196 MID 11 95
408 POST 18 60
184 POST 10 99
I separated out the data into three different data frames according to time level. So I have a SEH.pre, an SEH.mid and an SEH.post. I then do a describe and I know I have uneven groups of pre, mid, post. So, I want to random sample out pre, mid, post groups to be an even size. For example, I have the SEH.pre and SEH.mid group n sizes below:
> describe(SEH.pre)
vars n
Time.Level* 1 887
Demo.Age 2 883
SEH.Total 3 887
> describe(SEH.mid)
vars n
Time.Level* 1 894
Demo.Age 2 872
SEH.Total 3 894
So, now I run sample_n on the SEH.pre thinking that I can re-sample to an n of 860 across all columns. I run the following command:
SEH.pre2 <- sample_n(SEH.pre, 860, replace = FALSE)
And then I describe and the Demo.Age is less than the rest:
> describe(SEH.pre2)
vars n ...
Time.Level* 1 860
Demo.Age 2 856
SEH.Total 3 860
I feel like a big idiot but I cannot figure out why this is. I have tried it multiple times and Demo.Age varies from 856 to 859, but is never 860. I want all three columns to be 860. How do I do this? And why am I mis-thinking that sample_n should create even groups out of uneven?
I am a beginner with R . My data looks like this:
id count date
1 210 2009.01
2 400 2009.02
3 463 2009.03
4 465 2009.04
5 509 2009.05
6 861 2009.06
7 872 2009.07
8 886 2009.08
9 725 2009.09
10 687 2009.10
11 762 2009.11
12 748 2009.12
13 678 2010.01
14 699 2010.02
15 860 2010.03
16 708 2010.04
17 709 2010.05
18 770 2010.06
19 784 2010.07
20 694 2010.08
21 669 2010.09
22 689 2010.10
23 568 2010.11
24 584 2010.12
25 592 2011.01
26 548 2011.02
27 683 2011.03
28 675 2011.04
29 824 2011.05
30 637 2011.06
31 700 2011.07
32 724 2011.08
33 629 2011.09
34 446 2011.10
35 458 2011.11
36 421 2011.12
37 459 2012.01
38 256 2012.02
39 341 2012.03
40 284 2012.04
41 321 2012.05
42 404 2012.06
43 418 2012.07
44 520 2012.08
45 546 2012.09
46 548 2012.10
47 781 2012.11
48 704 2012.12
49 765 2013.01
50 571 2013.02
51 371 2013.03
I would like to make a bar graph like graph that shows how much what is the count for each date (dates in format of Month-Y, Jan-2009 for instance). I have two issues:
1- I cannot find a good format for a bar-char like graph like that
2- I want all of my data-points to be present in X axis(date), while R aggregates it to each year only (so I inly have four data-points there). Below is the current command that I am using:
plot(df$date,df$domain_count,col="red",type="h")
and my current plot is like this:
Ok, I see some issues in your original data. May I suggest the following:
Add the days in your date column
df$date=paste(df$date,'.01',sep='')
Convert the date column to be of date type:
df$date=as.Date(df$date,format='%Y.%m.%d')
Plot the data again:
plot(df$date,df$domain_count,col="red",type="h")
Also, may I add one more suggestion, have you used ggplot for ploting chart? I think you will find it much easier and resulting in better looking charts. Your example could be visualized like this:
library(ggplot2) #if you don't have the package, run install.packages('ggplot2')
ggplot(df,aes(date, count))+geom_bar(stat='identity')+labs(x="Date", y="Count")
First, you should transform your date column in a real date:
library(plyr) # for mutate
d <- mutate(d, month = as.numeric(gsub("[0-9]*\\.([0-9]*)", "\\1", as.character(date))),
year = as.numeric(gsub("([0-9]*)\\.[0-9]*", "\\1", as.character(date))),
Date = ISOdate(year, month, 1))
Then, you could use ggplot to create a decent barchart:
library(ggplot2)
ggplot(d, aes(x = Date, y = count)) + geom_bar(fill = "red", stat = "identity")
You can also use basic R to create a barchart, which is however less nice:
dd <- setNames(d$count, format(d$Date, "%m-%Y"))
barplot(dd)
The former plot shows you the "holes" in your data, i.e. month where there is no count, while for the latter it is even wuite difficult to see which bar corresponds to which month (this could however be tweaked I assume).
Hope that helps.
I have a data frame having 20 columns. I need to filter / remove noise from one column. After filtering using convolve function I get a new vector of values. Many values in the original column become NA due to filtering process. The problem is that I need the whole table (for later analysis) with only those rows where the filtered column has values but I can't bind the filtered column to original table as the number of rows for both are different. Let me illustrate using the 'age' column in 'Orange' data set in R:
> head(Orange)
Tree age circumference
1 1 118 30
2 1 484 58
3 1 664 87
4 1 1004 115
5 1 1231 120
6 1 1372 142
Convolve filter used
smooth <- function (x, D, delta){
z <- exp(-abs(-D:D/delta))
r <- convolve (x, z, type='filter')/convolve(rep(1, length(x)),z,type='filter')
r <- head(tail(r, -D), -D)
r
}
Filtering the 'age' column
age2 <- smooth(Orange$age, 5,10)
data.frame(age2)
The number of rows for age column and age2 column are 35 and 15 respectively. The original dataset has 2 more columns and I like to work with them also. Now, I only need 15 rows of each column corresponding to the 15 rows of age2 column. The filter here removed first and last ten values from age column. How can I apply the filter in a way that I get truncated dataset with all columns and filtered rows?
You would need to figure out how the variables line up. If you can add NA's to age2 and then do Orange$age2 <- age2 followed by na.omit(Orange) you should have what you want. Or, equivalently, perhaps this is what you are looking for?
df <- tail(head(Orange, -10), -10) # chop off the first and last 10 observations
df$age2 <- age2
df
Tree age circumference age2
11 2 1004 156 915.1678
12 2 1231 172 876.1048
13 2 1372 203 841.3156
14 2 1582 203 911.0914
15 3 118 30 948.2045
16 3 484 51 1008.0198
17 3 664 75 955.0961
18 3 1004 108 915.1678
19 3 1231 115 876.1048
20 3 1372 139 841.3156
21 3 1582 140 911.0914
22 4 118 32 948.2045
23 4 484 62 1008.0198
24 4 664 112 955.0961
25 4 1004 167 915.1678
Edit: If you know the first and last x observations will be removed then the following works:
x <- 2
df <- tail(head(Orange, -x), -x) # chop off the first and last x observations
df$age2 <- age2