I have a gridded data as a data-frame that has daily temperatures (in K) for 30 years. I need to extract data for days that matches another data-frame and keep the first and second columns (lon and lat).
Data example:
gridded data from which I need to remove days that do not match days in the second data (df2$Dates)
>head(Daily.df)
lon lat 1991-05-01 1991-05-02 1991-05-03 1991-05-04 1991-05-05 1991-05-06 1991-05-07 1991-05-08 1991-05-09
1 5.000 60 278.2488 280.1225 280.3909 279.4138 276.6809 276.2085 276.6250 277.7930 276.9693
2 5.125 60 278.2514 280.1049 280.3789 279.4395 276.7141 276.2467 276.6571 277.8264 277.0225
3 5.250 60 278.2529 280.0871 280.3648 279.4634 276.7437 276.2849 276.6918 277.8608 277.0740
4 5.375 60 278.2537 280.0687 280.3488 279.4858 276.7691 276.3238 276.7289 277.8960 277.1232
5 5.500 60 278.2537 280.0493 280.3319 279.5066 276.7909 276.3633 276.7688 277.9313 277.1701
6 5.625 60 278.2539 280.0294 280.3143 279.5264 276.8090 276.4042 276.8111 277.9666 277.2147
1991-05-10 1991-05-11 1991-05-12 1991-05-13 1991-05-14 1991-05-15 1991-05-16 1991-05-17 1991-05-18 1991-05-19
1 276.9616 277.3436 273.3149 274.4931 274.6967 275.6298 272.2511 271.5413 271.7289 271.7964
2 276.9689 277.2988 273.3689 274.5399 274.6801 275.6307 272.2214 271.4445 271.6410 271.7023
3 276.9720 277.2533 273.4225 274.5811 274.6646 275.6241 272.1858 271.3391 271.5424 271.5989
4 276.9716 277.2080 273.4726 274.6146 274.6507 275.6109 272.1456 271.2274 271.4340 271.4872
5 276.9689 277.1632 273.5163 274.6382 274.6380 275.5917 272.1022 271.1121 271.3168 271.3693
6 276.9645 277.1190 273.5507 274.6501 274.6263 275.5672 272.0571 270.9955 271.1919 271.2469
1991-05-20 1991-05-21 1991-05-22 1991-05-23 1991-05-24 1991-05-25 1991-05-26 1991-05-27 1991-05-28 1991-05-29
1 272.2633 268.0039 268.5981 269.4139 267.7836 265.8771 263.5669 266.1666 269.7285 272.5083
2 272.2543 268.0218 268.5847 269.4107 267.7886 265.8743 263.5125 266.1031 269.6471 272.4676
3 272.2434 268.0369 268.5716 269.4089 267.7910 265.8669 263.4592 266.0332 269.5697 272.4217
4 272.2308 268.0507 268.5597 269.4090 267.7925 265.8559 263.4066 265.9581 269.4987 272.3714
5 272.2164 268.0642 268.5505 269.4112 267.7936 265.8425 263.3546 265.8797 269.4355 272.3175
6 272.2005 268.0793 268.5451 269.4154 267.7962 265.8276 263.3039 265.7997 269.3818 272.2614
1991-05-30 1991-05-31 1991-06-01 1991-06-02 1991-06-03 1991-06-04 1991-06-05 1991-06-06 1991-06-07 1991-06-08
1 274.2950 273.4715 274.5197 274.7548 273.8259 272.4433 274.1811 274.4135 274.3999 276.0327
2 274.2205 273.4638 274.5292 274.8316 273.8658 272.4700 274.1992 274.4426 274.4650 276.0698
3 274.1421 273.4549 274.5373 274.9027 273.9028 272.4980 274.2160 274.4781 274.5309 276.1012
4 274.0609 273.4452 274.5438 274.9665 273.9365 272.5273 274.2322 274.5211 274.5969 276.1255
5 273.9784 273.4353 274.5482 275.0216 273.9660 272.5576 274.2481 274.5725 274.6617 276.1417
6 273.8960 273.4253 274.5508 275.0668 273.9912 272.5887 274.2649 274.6334 274.7239 276.1487
1991-06-09 1991-06-10 1991-06-11 1991-06-12 1991-06-13 1991-06-14 1991-06-15 1991-06-16 1991-06-17 1991-06-18
1 276.5216 277.1812 277.8093 278.3013 278.5323 278.5403 277.9563 278.3461 275.8296 273.8277
2 276.5531 277.1925 277.8261 278.3409 278.4956 278.5317 277.9148 278.3234 275.8167 273.8302
3 276.5861 277.2065 277.8457 278.3748 278.4503 278.5181 277.8654 278.2939 275.8057 273.8358
4 276.6204 277.2239 277.8684 278.4029 278.3988 278.4996 277.8080 278.2583 275.7966 273.8427
5 276.6564 277.2466 277.8945 278.4253 278.3423 278.4759 277.7429 278.2171 275.7888 273.8504
6 276.6938 277.2753 277.9242 278.4414 278.2834 278.4472 277.6715 278.1714 275.7819 273.8570
1991-06-19 1991-06-20 1991-06-21 1991-06-22 1991-06-23 1991-06-24 1991-06-25 1991-06-26 1991-06-27 1991-06-28
1 275.1738 274.6805 275.6100 274.8936 273.5818 273.2099 273.1788 271.2747 273.2458 276.9931
2 275.1808 274.7123 275.7043 274.9494 273.5861 273.1770 273.2280 271.2435 273.2662 276.9822
3 275.1859 274.7478 275.7993 275.0009 273.5956 273.1439 273.2730 271.2133 273.2803 276.9678
4 275.1891 274.7879 275.8941 275.0467 273.6107 273.1106 273.3130 271.1840 273.2886 276.9502
5 275.1902 274.8337 275.9870 275.0857 273.6318 273.0777 273.3472 271.1556 273.2918 276.9307
6 275.1891 274.8864 276.0776 275.1168 273.6589 273.0454 273.3752 271.1285 273.2905 276.9101
1991-06-29 1991-06-30
1 272.0784 273.5677
2 272.0577 273.5973
3 272.0339 273.6237
4 272.0075 273.6476
5 271.9794 273.6701
6 271.9500 273.6925
Second data I'm using for extracting (using Dates variable)
>head(df2)
Dates Temp Wind.S Wind.D
1 5/1/1991 18 4 238
2 5/2/1991 18 8 93
3 5/4/1991 22 8 229
4 5/6/1991 21 4 81
5 5/7/1991 21 8 192
6 5/9/1991 17 8 32
7 5/13/1991 22 8 229
8 5/18/1991 21 4 81
9 6/2/1991 21 8 192
10 6/7/1991 17 8 32
The header of the final data I'm looking for is something like this:
>head(df3)
lon lat 1991-05-01 1991-05-02 1991-05-04 1991-05-06 1991-05-09 1991-05-13
Example data following the format of yours
Daily.df <- data.frame(lon=1:5,lat=1:5,A=1:5,B=1:5,C=1:5,D=1:5)
colnames(Daily.df) <- c("lon","lat","1991-05-01","1991-05-02","1991-05-03","1991-05-04")
lon lat 1991-05-01 1991-05-02 1991-05-03 1991-05-04
1 1 1 1 1 1 1
2 2 2 2 2 2 2
3 3 3 3 3 3 3
4 4 4 4 4 4 4
5 5 5 5 5 5 5
df2 <- data.frame(Dates = c("5/1/1991","5/2/1991","5/4/1991"))
Dates
1 5/1/1991
2 5/2/1991
3 5/4/1991
Using lubridate to convert df2$Dates into the right format, make a vector of the column names you want to keep (thesedates) including lon and lat. Then use select_at to keep those columns.
library(lubridate)
library(dplyr)
thesedates <- c("lon","lat",as.character(mdy(df2$Dates)))
new.df <- Daily.df %>%
select_at(vars(thesedates))
Output
lon lat 1991-05-01 1991-05-02 1991-05-04
1 1 1 1 1 1
2 2 2 2 2 2
3 3 3 3 3 3
4 4 4 4 4 4
5 5 5 5 5 5
If you want to have a long data set to match, I would think you need to first convert the dates in df2 into the proper format and then wrangle the data into wide format.
Step 1 - Convert dates into correct format
df2$Dates <- as.Date(df2$Dates, format = "%m/%d/%Y")
Step 2 - convert to wide format
library(tidyr)
spread(df2, Dates, data)
Related
This question already has answers here:
For each row return the column name of the largest value
(10 answers)
Closed 12 months ago.
From a data frame in this structure
1 2 3
0.2393876 0.3388785 0.4217339
0.2984463 0.3948705 0.3066832
0.3042033 0.3321453 0.3636514
0.2290906 0.2796392 0.4912702
0.1971198 0.3273682 0.4755120
0.2328518 0.3991225 0.3680256
0.3031909 0.3790213 0.3177878
0.2734835 0.3233044 0.4032121
0.3190916 0.3256503 0.3552581
0.2600000 0.3281130 0.4118871
How is it possible to compare the value in the three columns, find the highest and add the name of the column in a new variable. Example output:
1 2 3 highestcolumn
0.2393876 0.3388785 0.4217339 3
0.2984463 0.3948705 0.3066832 2
0.3042033 0.3321453 0.3636514 3
Use which.max and apply on each row.
df$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$highestcolumn <- max.col(df, "first")
-output
> df
1 2 3 highestcolumn
1 0.2393876 0.3388785 0.4217339 3
2 0.2984463 0.3948705 0.3066832 2
3 0.3042033 0.3321453 0.3636514 3
4 0.2290906 0.2796392 0.4912702 3
5 0.1971198 0.3273682 0.4755120 3
6 0.2328518 0.3991225 0.3680256 2
7 0.3031909 0.3790213 0.3177878 2
8 0.2734835 0.3233044 0.4032121 3
9 0.3190916 0.3256503 0.3552581 3
10 0.2600000 0.3281130 0.4118871 3
I am not to proficient with the apply functions, or with R. But I know I overuse for loops which makes my code slow. How can the following code be sped up with apply functions, or in any other way?
sum_store = NULL
for (col in 1:ncol(cazy_fams)){ # for each column in cazy_fams (so for each master family eg. GH, AA ect...)
for (row in 1:nrow(cazy_fams)){ # for each row in cazy fams (so the specific family number e.g GH1 AA7 ect...)
# Isolating the row that pertains to the current cazy family being looked at for every dataframe in the list
filt_fam = lapply(family_summary, function(sample){
sample[as.character(sample$Family) %in% paste(colnames(cazy_fams[col]),cazy_fams[row,col], sep = ""),]
})
row_cat = do.call(rbind, filt_fam) # concatinating the lapply list output int a dataframe
if (nrow(row_cat) > 0){
fam_sum = aggregate(proteins ~ Family, data=row_cat, FUN=sum) # collapsing the dataframe into one row and summing the proteins count
sum_store = rbind(sum_store, fam_sum) # storing the results for that family
} else if (grepl("NA", paste(colnames(cazy_fams[col]),cazy_fams[row,col], sep = "")) == FALSE) {
Family = paste(colnames(cazy_fams[col]),cazy_fams[row,col], sep = "")
proteins = 0
sum_store = rbind(sum_store, data.frame(Family, proteins))
} else {
next
}
}
}
family_summary is just a list of 18 two column dataframes that look like this:
Family proteins
CE0 2
CE1 9
CE4 15
CE7 1
CE9 1
CE14 10
GH0 5
GH1 1
GH3 4
GH4 1
GH8 1
GH9 2
GH13 2
GH15 5
GH17 1
with different cazy families.
cazy_fams is just a dataframe with each coulms being a cazy class (eg. GH, AA ect...) and ech row being a family number, all taken from the linked website:
GH GT PL CE AA CBM
1 1 1 1 1 1
2 2 2 2 2 2
3 3 3 3 3 3
4 4 4 4 4 4
5 5 5 5 5 5
6 6 6 6 6 6
7 7 7 7 7 7
8 8 8 8 8 8
9 9 9 9 9 9
10 10 10 10 10 10
11 11 11 11 11 11
12 12 12 12 12 12
13 13 13 13 13 13
14 14 14 14 14 14
15 15 15 15 15 15
The reason behind the else if (grepl("NA", paste(colnames(cazy_fams[col]),cazy_fams[row,col], sep = "")) == FALSE) statment is to deal with the fact not all classes have the same number of family so when looping over my dataframe I end up with some GHNA and AANA with NA on the end.
The output sum_store is this:
Family proteins
GH1 54
GH2 51
GH3 125
GH4 29
GH5 40
GH6 25
GH7 0
GH8 16
GH9 25
GH10 19
GH11 5
GH12 5
GH13 164
GH14 3
GH15 61
A dataframe with all listed cazy families and the total number of apperances across the family_summary list.
Please let me know if you need anything else to help answer my question.
I have to combine 5 files with the same structure and add a new variable to index the new data frame, but all 5 files are using the same ID.
I successfully combine them but I do not find how to index them. I have tried a few loops, but they were not giving me what I wanted.
# Combining files
path <- "D:/..."
filenames <- list.files(path)
t <- do.call("rbind", lapply(filenames, read.table, header = TRUE))
# Trying indexing with loops:
for (i in 1:length(t$ID){
t$ID2<-(t$ID+last(t$ID2))
}
I have 5 files, all of them with the same structure, and all of them using the same variable for identification, i.e.
file 1 would have:
ID: 1 1 1 2 2 2 3 3 3
And file 2 to 5 would have exactly the same IDs:
I would like to combine them into a single data frame so I would have this:
ID: 1 1 1 2 2 2 3 3 3 1 1 1 2 2 2 3 3 3 1 1 1....
and then name them differently. So I would have:
ID: 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6 7 7 7...
How's this? This code finds the largest ID of first (i) data.frame and then adds that to IDs of next (i+1) data.frame. It records (i+1) largest ID and uses that in the (i+2) data.frame.
For this to work, you will have to forego the first do.call(rbind, ...) in your code.
xy1 <- data.frame(id = rep(1:4, each = 4), matrix(runif(4*4 * 3), ncol = 3))
xy2 <- data.frame(id = rep(1:7, each = 3), matrix(runif(3*7 * 3), ncol = 3))
xy3 <- data.frame(id = rep(1:3, each = 5), matrix(runif(3*5 * 3), ncol = 3))
xy <- list(xy1, xy2, xy3)
# First find largest ID of the first data.frame.
maxid <- max(xy[[1]]$id)
# Add previous max to current ID.
for (i in 2:length(xy)) {
xy[[i]]$id <- maxid + xy[[i]]$id
maxid <- max(xy[[i]]$id) # calculates largest id to be used next
}
> do.call(rbind, xy)
id X1 X2 X3
1 1 0.881397055 0.113236016 0.58935016
2 1 0.205762300 0.216630633 0.04096480
3 1 0.307112552 0.005092413 0.97769030
4 1 0.457299727 0.329346925 0.09582600
5 2 0.007010529 0.089751397 0.69746047
6 2 0.014806573 0.432586138 0.44480438
7 2 0.534909561 0.108258153 0.82475185
8 2 0.313796157 0.749077837 0.38798818
9 3 0.643547518 0.237040912 0.18304776
10 3 0.725906336 0.186099719 0.61738806
11 3 0.506767958 0.646870554 0.27792817
12 3 0.303638439 0.082478410 0.52484137
13 4 0.360623223 0.182054933 0.48604454
14 4 0.804174231 0.427352128 0.70075198
15 4 0.211255624 0.673377745 0.77251727
16 4 0.474358562 0.430095921 0.03648586
17 5 0.731251361 0.635859860 0.90235962
18 5 0.689463703 0.931878683 0.12179179
19 5 0.256770523 0.413928661 0.89254294
20 6 0.358319709 0.393714347 0.53143877
21 6 0.241538687 0.811901018 0.91577045
22 6 0.445141806 0.015133252 0.70977512
23 7 0.179662683 0.574578297 0.09957555
24 7 0.279302309 0.351412534 0.40911867
25 7 0.826039704 0.852739191 0.58671811
26 8 0.822024888 0.061122387 0.12308001
27 8 0.676081285 0.005285565 0.32040908
28 8 0.302821623 0.511678250 0.14814015
29 9 0.966690845 0.221078055 0.72651928
30 9 0.070768391 0.726477379 0.70431920
31 9 0.178425952 0.223096153 0.41111805
32 10 0.952963096 0.209673890 0.73485060
33 10 0.905570765 0.290359419 0.69499805
34 10 0.976600565 0.448144677 0.36100322
35 11 0.458720466 0.636912805 0.04170255
36 11 0.953471285 0.533102906 0.63543974
37 11 0.574490192 0.975327747 0.94730912
38 12 0.878968237 0.956726315 0.04761167
39 12 0.379196322 0.720179957 0.98719308
40 12 0.217246809 0.066895905 0.44981063
41 12 0.309354927 0.048701078 0.24654953
42 12 0.011187546 0.833095978 0.94793368
43 13 0.590529610 0.240967648 0.42954908
44 13 0.525187039 0.739698883 0.72047067
45 13 0.223469798 0.338660741 0.21820068
46 13 0.359939747 0.831732199 0.27095365
47 13 0.672778236 0.327900275 0.04854854
48 14 0.202447020 0.911963711 0.18576047
49 14 0.858830035 0.003633945 0.25713498
50 14 0.784197766 0.527018979 0.30911792
51 14 0.942135786 0.256841256 0.76965498
52 14 0.488395595 0.716133306 0.89618736
I'd like to use the pmax command to create a new column. My code Looks like this:
Master <- Master %>%
mutate(RAM = pmax(RAM1, RAM2, RAM3, RAM4, RAM5, RAM6, RAM7, RAM8, RAM9, RAM10,
RAM11, RAM12, RAM13, RAM14, RAM15, RAM16, RAM17, RAM18,
RAM19, RAM20, RAM21, RAM22, RAM23, RAM24, RAM25, RAM26,
RAM27, RAM28, RAM29, RAM30, RAM31, RAM32, RAM33, RAM34,
RAM35, RAM36, RAM37, RAM38, RAM39, RAM40, RAM41, RAM42,
RAM43, RAM44, RAM45, RAM46, RAM47, RAM48, RAM49, RAM50,
RAM51, RAM52, RAM53, RAM54, RAM55, RAM56, RAM57, RAM58,
RAM59, RAM60, RAM61, RAM62, RAM63, RAM64, RAM65, RAM66,
RAM67, RAM68, RAM69, RAM70, RAM71, RAM72, RAM73, RAM74,
RAM75, RAM76, RAM77, RAM78, RAM79, RAM80, RAM81, RAM82,
RAM83, RAM84, RAM85, RAM86, RAM87, RAM88, RAM89, RAM90,
RAM91, RAM92, na.rm =T))
In my current data base, however, only the columns RAM1 to RAM8 exist. In this case, I want R to skip all the other columns mentioned in the Statement and to only use column RAM1 to RAM8 (it is okay if R displays an error message, but I don't want the program to interrupt running the code).
Any ideas how to do so?
Thanks!
One way to do this would be as follows:
Set up some data to make a reproducible example
set.seed(0)
Master <- data.frame(Other=100,RAM1=1:10, RAM2=1:10, RAM3=1:10, RAM4=1:10,
RAM5=1:10, RAM6=1:10, RAM7=1:10, RAM8=rnorm(10)+5)
Master[5,5] <- NA
Select required columns of the dataframe:
Master[colnames(Master) %in% paste0("RAM",1:92)]
Use do.call to run pmax using the selected columns as arguments, and adding the argument na.rm=TRUE
Master$RAM <- do.call(pmax, c(Master[colnames(Master) %in% paste0("RAM",1:92)], na.rm=TRUE))
Sample output:
Master
# Other RAM1 RAM2 RAM3 RAM4 RAM5 RAM6 RAM7 RAM8 RAM
#1 100 1 1 1 1 1 1 1 6.262954 6.262954
#2 100 2 2 2 2 2 2 2 4.673767 4.673767
#3 100 3 3 3 3 3 3 3 6.329799 6.329799
#4 100 4 4 4 4 4 4 4 6.272429 6.272429
#5 100 5 5 5 NA 5 5 5 5.414641 5.414641
#6 100 6 6 6 6 6 6 6 3.460050 6.000000
#7 100 7 7 7 7 7 7 7 4.071433 7.000000
#8 100 8 8 8 8 8 8 8 4.705280 8.000000
#9 100 9 9 9 9 9 9 9 4.994233 9.000000
#10 100 10 10 10 10 10 10 10 7.404653 10.000000
According to solution in Gather multiple sets of columns I am trying to gather multiple sets of columns. All columns which I would like to gather have year in the end of the sentece. column.name.2010, column.name.2011, column.name.2012, column.name2.2010, column.name2.2011, column.name2.2012 etc.
I tried
db %>%
gather(key, value, -c(1:6)) %>%
extract(key, c("question", "Year"), "([^\d{4}]+)") %>%
spread(question, value)
I would like to accomplish this:
Id Index.number Country.Name World.Rank.by.Assets Year column.name. column.name2.
1 a 3 6 2010 xxxx xxy
2 b 4 6 2011 xxyx xyy
3 c 5 8 2012 xxxs sada
Head of db
Id Index.number Country.Name World.Rank.by.Assets Total.Assets.th.EUR.2010 Total.Assets.th.EUR.2011
1 10931 10931 9 6 n.a. 1,965,283,000
2 10696 10696 9 9 n.a. 1,879,536,000
3 11948 11948 9 13 n.a. 1,723,608,000
4 11150 11150 9 19 n.a. 1,181,372,000
5 10708 10708 9 NA n.a. 1,138,395,000
Total.Assets.th.EUR.2012 Total.Assets.th.EUR.2013 Total.Assets.th.EUR.2014 Total.Assets.th.EUR.2015
1 1,907,200,000 1,810,522,000 2,077,759,000 1,994,193,000
2 1,783,220,000 1,688,264,000 1,762,714,000 1,698,859,000
3 1,617,429,000 1,518,811,000 1,589,044,000 1,529,294,000
4 1,250,889,000 1,214,193,000 1,308,138,000 1,334,391,000
5 1,147,521,000 1,123,520,000 1,223,298,000 1,166,535,000
Total.Assets.th.EUR.2016 Deposits...Short.term.funding.th.EUR.2010
1 2,076,959,000 n.a.
2 1,722,849,000 n.a.
3 1,524,232,000 n.a.
4 1,382,241,000 n.a.
5 1,235,240,000 n.a.
Deposits...Short.term.funding.th.EUR.2011 Deposits...Short.term.funding.th.EUR.2012
1 947,516,000 942,644,000
2 917,933,000 928,045,000
3 808,028,000 805,504,000
4 452,417,000 540,628,000
5 604,912,000 661,297,000
Deposits...Short.term.funding.th.EUR.2013 Deposits...Short.term.funding.th.EUR.2015
1 942,387,000 1,032,100,000
2 911,884,000 891,035,000
3 789,353,000 796,777,000
4 523,624,000 673,520,000
5 692,997,000 707,185,000
Deposits...Short.term.funding.th.EUR.2014 Deposits...Short.term.funding.th.EUR.2016
1 1,031,624,000 1,111,246,000
2 890,072,000 903,317,000
3 773,094,000 759,184,000
4 587,177,000 687,328,000
5 740,298,000 761,140,000
Loan.loss.res...Gross.loans...2010 Loan.loss.res...Gross.loans...2011 Loan.loss.res...Gross.loans...2012
1 n.a. 4,04 4,05
2 n.a. 3,55 3,16
3 n.a. 4,53 3,77
4 n.a. 4,21 4,26
5 n.a. 1,96 2,03