This may be a simple question, but I haven't been able to find any answer. Consider you have a dataframe with n columns with molecular features. In the last row of each column, a coefficient of variance is expressed.
Example data set:
a <- data.frame(matrix(runif(30),ncol=3))
b <- c(50.23,45.23,21)
a<-rbind(a,b)
X1 X2 X3
1 0.1097075 0.78584027 0.20925033
2 0.6081752 0.39669748 0.65559913
3 0.9912855 0.68462073 0.54741795
4 0.8543848 0.53776889 0.43789447
5 0.2579654 0.92188090 0.61292895
6 0.6203840 0.73152279 0.82866311
7 0.6643195 0.84953926 0.62192976
8 0.5760624 0.30949900 0.11032929
9 0.8888167 0.04530598 0.08089825
10 0.8926815 0.61736284 0.19834310
11 50.2300000 45.23000000 21.00000000
How do I subset so I only get the columns with CV>50 in the last row? So my new data.frame would be:
X1
1 0.1097075
2 0.6081752
3 0.9912855
4 0.8543848
5 0.2579654
6 0.6203840
7 0.6643195
8 0.5760624
9 0.8888167
10 0.8926815
11 50.230000
We can do
a[,a[nrow(a),]>50,drop=FALSE]
Related
This question already has answers here:
For each row return the column name of the largest value
(10 answers)
Closed 12 months ago.
From a data frame in this structure
1 2 3
0.2393876 0.3388785 0.4217339
0.2984463 0.3948705 0.3066832
0.3042033 0.3321453 0.3636514
0.2290906 0.2796392 0.4912702
0.1971198 0.3273682 0.4755120
0.2328518 0.3991225 0.3680256
0.3031909 0.3790213 0.3177878
0.2734835 0.3233044 0.4032121
0.3190916 0.3256503 0.3552581
0.2600000 0.3281130 0.4118871
How is it possible to compare the value in the three columns, find the highest and add the name of the column in a new variable. Example output:
1 2 3 highestcolumn
0.2393876 0.3388785 0.4217339 3
0.2984463 0.3948705 0.3066832 2
0.3042033 0.3321453 0.3636514 3
Use which.max and apply on each row.
df$highestcolumn <- apply(df, 1, which.max)
> df
X1 X2 X3 mean
1 0.2393876 0.3388785 0.4217339 3
2 0.2984463 0.3948705 0.3066832 2
3 0.3042033 0.3321453 0.3636514 3
4 0.2290906 0.2796392 0.4912702 3
5 0.1971198 0.3273682 0.4755120 3
6 0.2328518 0.3991225 0.3680256 2
7 0.3031909 0.3790213 0.3177878 2
8 0.2734835 0.3233044 0.4032121 3
9 0.3190916 0.3256503 0.3552581 3
10 0.2600000 0.3281130 0.4118871 3
We may use max.col
df$highestcolumn <- max.col(df, "first")
-output
> df
1 2 3 highestcolumn
1 0.2393876 0.3388785 0.4217339 3
2 0.2984463 0.3948705 0.3066832 2
3 0.3042033 0.3321453 0.3636514 3
4 0.2290906 0.2796392 0.4912702 3
5 0.1971198 0.3273682 0.4755120 3
6 0.2328518 0.3991225 0.3680256 2
7 0.3031909 0.3790213 0.3177878 2
8 0.2734835 0.3233044 0.4032121 3
9 0.3190916 0.3256503 0.3552581 3
10 0.2600000 0.3281130 0.4118871 3
I'm currently reading "Practical Statistics for Data Scientists" and following along in R as they demonstrate some code. There is one chunk of code I'm particularly struggling to follow the logic of and was hoping someone could help. The code in question is creating a dataframe with 1000 rows where each observation is the mean of 5 randomly drawn income values from the dataframe loans_income. However, I'm getting confused about the logic of the code as it is fairly complicated with a tapply() function and nested rep() statements.
The code to create the dataframe in question is as follows:
samp_mean_5 <- data.frame(income = tapply(sample(loans_income$income,1000*5),
rep(1:1000,rep(5,1000)),
FUN = mean),
type='mean_of_5')
In particular, I'm confused about the nested rep() statements and the 1000*5 portion of the sample() function. Any help understanding the logic of the code would be greatly appreciated!
For reference, the original dataset loans_income simply has a single column of 50,000 income values.
You have 50,000 loans_income in a single vector. Let's break your code down:
tapply(sample(loans_income$income,1000*5),
rep(1:1000,rep(5,1000)),
FUN = mean)
I will replace 1000 with 10 and income with random numbers, so it's easier to explain. I also set set.seed(1) so the result can be reproduced.
sample(loans_income$income,1000*5)
We 50 random incomes from your vector without replacement. They are (temporarily) put into a vector of length 50, so the output looks like this:
> sample(runif(50000),10*5)
[1] 0.73283101 0.60329970 0.29871173 0.12637654 0.48434952 0.01058067 0.32337850
[8] 0.46873561 0.72334215 0.88515494 0.44036341 0.81386225 0.38118213 0.80978822
[15] 0.38291273 0.79795343 0.23622492 0.21318431 0.59325586 0.78340477 0.25623138
[22] 0.64621658 0.80041393 0.68511759 0.21880083 0.77455662 0.05307712 0.60320912
[29] 0.13191926 0.20816298 0.71600799 0.70328349 0.44408218 0.32696205 0.67845445
[36] 0.64438336 0.13241312 0.86589561 0.01109727 0.52627095 0.39207860 0.54643661
[43] 0.57137320 0.52743012 0.96631114 0.47151170 0.84099503 0.16511902 0.07546454
[50] 0.85970500
rep(1:1000,rep(5,1000))
Now we are creating an indexing vector of length 50:
> rep(1:10,rep(5,10))
[1] 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 4 4 4 4 4 5 5 5 5 5 6 6 6
[29] 6 6 7 7 7 7 7 8 8 8 8 8 9 9 9 9 9 10 10 10 10 10
Those indices "group" the samples from step 1. So basically this vector tells R that the first 5 entries of your "sample vector" belong together (index 1), the next 5 entries belong together (index 2) and so on.
FUN = mean
Just apply the mean-function on the data.
tapply
So tapply takes the sampled data (sample-part) and groups them by the second argument (the rep()-part) and applies the mean-function on each group.
If you are familiar with data.frames and the dplyr package, take a look at this (only the first 10 rows are displayed):
set.seed(1)
df <- data.frame(income=sample(runif(5000),10*5), index=rep(1:10,rep(5,10)))
income index
1 0.42585569 1
2 0.16931091 1
3 0.48127444 1
4 0.68357403 1
5 0.99374923 1
6 0.53227877 2
7 0.07109499 2
8 0.20754511 2
9 0.35839481 2
10 0.95615917 2
I attached the an index to the random numbers (your income). Now we calculate the mean per group:
df %>%
group_by(index) %>%
summarise(mean=mean(income))
which gives us
# A tibble: 10 x 2
index mean
<int> <dbl>
1 1 0.551
2 2 0.425
3 3 0.827
4 4 0.391
5 5 0.590
6 6 0.373
7 7 0.514
8 8 0.451
9 9 0.566
10 10 0.435
Compare it to
set.seed(1)
tapply(sample(runif(5000),10*5),
rep(1:10,rep(5,10)),
mean)
which yields basically the same result:
1 2 3 4 5 6 7 8 9
0.5507529 0.4250946 0.8273149 0.3905850 0.5902823 0.3730092 0.5143829 0.4512932 0.5658460
10
0.4352546
I have a dataset of Ages for the customer and I wanted to make a frequency distribution by 9 years of a gap of age.
Ages=c(83,51,66,61,82,65,54,56,92,60,65,87,68,64,51,
70,75,66,74,68,44,55,78,69,98,67,82,77,79,62,38,88,76,99,
84,47,60,42,66,74,91,71,83,80,68,65,51,56,73,55)
My desired outcome would be similar to below-shared table, variable names can be differed(as you wish)
Could I use binCounts code into it ? if yes could you help me out using the code as not sure of bx and idxs in this code?
binCounts(x, idxs = NULL, bx, right = FALSE) ??
Age Count
38-46 3
47-55 7
56-64 7
65-73 14
74-82 10
83-91 6
92-100 3
Much Appreciated!
I don't know about the binCounts or even the package it is in but i have a bare r function:
data.frame(table(cut(Ages,0:7*9+37)))
Var1 Freq
1 (37,46] 3
2 (46,55] 7
3 (55,64] 7
4 (64,73] 14
5 (73,82] 10
6 (82,91] 6
7 (91,100] 3
To exactly duplicate your results:
lowerlimit=c(37,46,55,64,73,82,91,101)
Labels=paste(head(lowerlimit,-1)+1,lowerlimit[-1],sep="-")#I add one to have 38 47 etc
group=cut(Ages,lowerlimit,Labels)#Determine which group the ages belong to
tab=table(group)#Form a frequency table
as.data.frame(tab)# transform the table into a dataframe
group Freq
1 38-46 3
2 47-55 7
3 56-64 7
4 65-73 14
5 74-82 10
6 83-91 6
7 92-100 3
All this can be combined as:
data.frame(table(cut(Ages,s<-0:7*9+37,paste(head(s+1,-1),s[-1],sep="-"))))
In this hypothetical scenario, I have performed 5 different analyses on 13 chemicals, resulting in a score assigned to each chemical within each analysis. I have created a table as follows:
---- Analysis1 Analysis2 Analysis3 Analysis4 Analysis5
Chem_1 3.524797844 4.477695034 4.524797844 4.524797844 4.096698498
Chem_2 2.827511555 3.827511555 3.248136118 3.827511555 3.234398548
Chem_3 2.682144761 3.474646298 3.017780505 3.682144761 3.236152242
Chem_4 2.134137304 2.596921333 2.95181339 2.649076603 2.472875191
Chem_5 2.367736454 3.027814219 2.743137896 3.271122346 2.796607809
Chem_6 2.293110565 2.917318708 2.724156207 3.293110565 2.530967343
Chem_7 2.475709113 3.105794018 2.708222528 3.475709113 3.088819908
Chem_8 2.013451822 2.259454085 2.683273938 2.723554966 2.400976121
Chem_9 2.345123123 3.050074893 2.682845391 3.291851228 2.700844104
Chem_10 2.327658894 2.848729452 2.580415233 3.327658894 2.881490893
Chem_11 2.411243882 2.98131398 2.554456095 3.411243882 3.109205453
Chem_12 2.340778276 2.576860244 2.549707035 3.340778276 3.236545826
Chem_13 2.394698249 2.90682524 2.542599327 3.394698249 3.12936843
I would like to create columns corresponding to each analysis which contain the rank position for each chemical. For instance, under Analysis1,Chem_1 would have value "1", Chem_2 would have value "2", Chem_3 would have value "4", Chem_7 would have value "4", Chem_11 would have value "5", and so on.
We can use dense_rank from dplyr
library(dplyr)
df %>%
mutate_each(funs(dense_rank(-.)))
In base R, we can do
df[] <- lapply(-df, rank, ties.method="min")
In data.table, we can use
library(data.table)
setDT(df)[, lapply(-.SD, frank, ties.method="dense")]
To avoid the copies from multiplying with -, as #Arun mentioned in the comments
lapply(.SD, frankv, order=-1L, ties.method="dense")
You can also do this in base R:
cbind("..." = df[,1], data.frame(do.call(cbind,
lapply(df[,-1], order, decreasing = T))))
... Analysis1 Analysis2 Analysis3 Analysis4 Analysis5
1 Chem_1 1 1 1 1 1
2 Chem_2 2 2 2 2 12
3 Chem_3 3 3 3 3 3
4 Chem_4 7 7 4 7 2
5 Chem_5 11 9 5 11 13
6 Chem_6 13 5 6 13 11
7 Chem_7 5 11 7 12 7
8 Chem_8 9 6 8 10 10
9 Chem_9 12 13 9 6 5
10 Chem_10 10 10 10 9 9
11 Chem_11 6 4 11 5 6
12 Chem_12 4 12 12 8 4
13 Chem_13 8 8 13 4 8
If I'm not mistaken, you want to have the column-wise rank of your table. Here is my solution:
m=data.matrix(df) # converts data frame to matrix, convert your data to matrix accordingly
apply(m, 2, function(c) rank(c)) # increasingly
apply(m, 2, function(c) rank(-c)) # decreasingly
However, I believe you could solve it by yourself with the help of the answers to this question
Get rank of matrix entries?
Suppose I have a data.frame a (each observation is in a row) and calculated the distance matrix. the question is, is there any function that will give the distances of observation 5 to all the other observations.
> a=data.frame(A=rnorm(10), B=rnorm(10), C=rnorm(10))
> b=dist(a)
> b
1 2 3 4 5 6 7 8 9
2 1.6118634
3 0.4891468 1.3382692
4 1.2002947 1.7516061 0.9160975
5 1.8128570 0.3197837 1.5192406 1.7709168
6 0.7280433 1.2628696 0.4063128 1.2411639 1.4971098
7 1.7616767 0.7400666 1.4512844 1.4355922 0.5168996 1.5524980
8 3.1033274 3.3739578 2.7297046 2.2281075 3.3693333 2.7738859 3.1216145
9 2.0916857 1.6749526 1.6717408 2.0293415 1.8196557 1.3704288 1.9824870 2.4013682
10 1.5949320 1.7309838 1.1680365 0.6331770 1.7255615 1.3234977 1.4333926 1.7798153 1.6126823
Just convert it to a matrix:
as.matrix(b)[5,]
Check out the pdist package. It basically returns a single row of what as.matrix(dist(x)) would return... so you don't have to calculate everything.
http://cran.r-project.org/web/packages/pdist/index.html