Reconstruct dataframe with dates as date intervals in R - r

I have a dataset basically looks like that, giving which campaigns are active for each household with given start and end dates of respective campaigns:
campaign_id household_id campaign_type start_date end_date
1 26 1 Type B 2016-12-28 2017-02-19
2 8 1 Type A 2017-05-08 2017-06-25
3 12 1 Type B 2017-07-12 2017-08-13
4 13 1 Type A 2017-08-08 2017-09-24
5 18 1 Type A 2017-10-30 2017-12-24
6 20 1 Type C 2017-11-27 2018-02-05
7 22 1 Type B 2017-12-06 2018-01-07
8 23 1 Type B 2017-12-28 2018-02-04
And I create a new dataframe with given structure, which will show which campaigns are active for given household in a given time (having all the campaign numbers as columns, i have omitted the rest while putting here):
household_id date campaign1 campaign2 campaign3 campaign4
1 1 2016-11-14 NA NA NA NA
2 1 2016-12-06 NA NA NA NA
3 1 2016-12-28 NA NA NA NA
4 1 2017-02-08 NA NA NA NA
5 1 2017-03-03 NA NA NA NA
6 1 2017-03-08 NA NA NA NA
7 1 2017-03-13 NA NA NA NA
8 1 2017-03-29 NA NA NA NA
9 1 2017-04-03 NA NA NA NA
10 1 2017-04-19 NA NA NA NA
What I want to do is assigning the active promotions in the given dates as rows in the second dataframe. For example if household_id 1 is having campaign 2 running in 2016-11-14 but no other campaigns, then it will look like this:
household_id date campaign1 campaign2 campaign3 campaign4
1 1 2016-11-14 0 1 0 0
How can i manage this construction, should I use for loops in the initial dataframe and assign to second one in each loop, or there is a better and faster way? Thanks in advance.

Related

r data.table : lagging a date variable [duplicate]

This question already has answers here:
How to create a lag variable within each group?
(5 answers)
Closed 2 years ago.
I have data that looks similar to the following except with hundreds of IDs and thousands of observations:
ID date measles
1 2008-09-12 1
1 2008-10-25 NA
1 2009-01-12 1
1 2009-03-12 NA
1 2009-05-12 1
2 2010-05-17 NA
2 2010-06-12 NA
2 2010-07-02 1
2 2010-08-13 NA
I want to create a variable that will store the previous date for each pid like the following:
ID date measles previous_date
1 2008-09-12 1 NA
1 2008-10-25 NA 2008-09-12
1 2009-01-12 1 2008-10-25
1 2009-03-12 NA 2009-01-12
1 2009-05-12 1 2009-03-12
2 2010-05-17 NA NA
2 2010-06-12 NA 2010-05-17
2 2010-07-02 1 2010-06-12
2 2010-08-13 NA 2010-07-02
This should be an extremely easy task, but I have been unsuccessful at getting a lag variable to work properly. I have tried a few methods, such as the following:
dt[, previous_date:=c(NA, current_date[-.N]), by=c("ID")]
dt[,previous_date:=current_date-shift(current_date,1,type="lag"),by=ID]
The code samples above either produce sporadic numbers in the previous_date variable or produce all NAs. I'm not sure why this is? Is it because I'm using a date variable as opposed to an integer?
Is there a better way to accomplish this task that would work for a date variable?

We can just use shift on the 'date' column grouped by 'ID'. By default the type is lag
library(data.table)
dt[, previous_date := shift(date), ID]
dt
# ID date measles previous_date
#1: 1 2008-09-12 1 <NA>
#2: 1 2008-10-25 NA 2008-09-12
#3: 1 2009-01-12 1 2008-10-25
#4: 1 2009-03-12 NA 2009-01-12
#5: 1 2009-05-12 1 2009-03-12
#6: 2 2010-05-17 NA <NA>
#7: 2 2010-06-12 NA 2010-05-17
#8: 2 2010-07-02 1 2010-06-12
#9: 2 2010-08-13 NA 2010-07-02

Create a column that assigns value to a row in a dataframe based on an event in another row

I have a dataframe that is structured like the following:
example <- data.frame(id = c(1,1,1,1,1,1,1,2,2,2,2,2),
event = c("email","email","email","draw","email","email","draw","email","email","email","email","draw"),
date = c("2020-03-01","2020-06-01","2020-07-15","2020-07-28","2020-08-07","2020-09-01","2020-09-15","2020-05-22","2020-06-15","2020-07-13","2020-07-15","2020-07-31"),
amount = c(NA,NA,NA,10000,NA,NA,1500,NA,NA,NA,NA,2200))
This is a simplified version of the dataframe. I am trying to create a column that will assign a 1 to the last email before the draw event and a column that will have the amount drawn on the same row as the email. The desired dataframe would look like the following:
desiredResult <- data.frame(id = c(1,1,1,1,1,1,1,2,2,2,2,2),
event = c("email","email","email","draw","email","email","draw","email","email","email","email","draw"),
date = c("2020-03-01","2020-06-01","2020-07-15","2020-07-28","2020-08-07","2020-09-01","2020-09-15","2020-05-22","2020-06-15","2020-07-13","2020-07-15","2020-07-31"),
amount = c(NA,NA,NA,10000,NA,NA,1500,NA,NA,NA,NA,2200),
EmailBeforeDrawFlag = c(NA,NA,1,NA,NA,1,NA,NA,NA,NA,1,NA),
EmailBeforeDrawAmount = c(NA,NA,10000,NA,NA,1500,NA,NA,NA,NA,2200,NA))

Here is the dplyr solution. When you create new columns, you want to use if_else() in the definition of EmailBeforeDrawFlag to check a condition, and the lead function to look in the previous row for event. EmailBeforeDrawAmount is juts lead(amount).
example %>%
mutate(EmailBeforeDrawFlag = if_else(lead(event) == "draw", 1, NA_real_ ),
EmailBeforeDrawAmount = lead(amount))
id event date amount EmailBeforeDrawFlag EmailBeforeDrawAmount
1 1 email 2020-03-01 NA NA NA
2 1 email 2020-06-01 NA NA NA
3 1 email 2020-07-15 NA 1 10000
4 1 draw 2020-07-28 10000 NA NA
5 1 email 2020-08-07 NA NA NA
6 1 email 2020-09-01 NA 1 1500
7 1 draw 2020-09-15 1500 NA NA
8 2 email 2020-05-22 NA NA NA
9 2 email 2020-06-15 NA NA NA
10 2 email 2020-07-13 NA NA NA
11 2 email 2020-07-15 NA 1 2200
12 2 draw 2020-07-31 2200 NA NA

We could also make use of NA^ to create the column on the lead
library(dplyr)
example %>%
mutate(EmailBeforeDrawFlag = NA^(lead(event != 'draw')),
EmailBeforeDrawAmount = lead(amount))
-output
# id event date amount EmailBeforeDrawFlag EmailBeforeDrawAmount
#1 1 email 2020-03-01 NA NA NA
#2 1 email 2020-06-01 NA NA NA
#3 1 email 2020-07-15 NA 1 10000
#4 1 draw 2020-07-28 10000 NA NA
#5 1 email 2020-08-07 NA NA NA
#6 1 email 2020-09-01 NA 1 1500
#7 1 draw 2020-09-15 1500 NA NA
#8 2 email 2020-05-22 NA NA NA
#9 2 email 2020-06-15 NA NA NA
#10 2 email 2020-07-13 NA NA NA
#11 2 email 2020-07-15 NA 1 2200
#12 2 draw 2020-07-31 2200 NA NA

Dividing the data in multiple columns to 8 values logically in R

I have the data as following. Each column starting from 1.07m to 11.82m represents the depth and the values corresponds to the temperature. I am interested in reducing the datasets into 8 sets (8 distinct water depths). While doing so I would like to use the averaging method. For example in row1 of my data starts from column x1.07m to x2.82m (x2.82m because all the values beyond that point are NA). I would like to create a separate data frame with data and 8 columns (layer1, layer2, layer3, layer4, layer5, layer6, layer7, layer8). Layer1 value should start from 1.07m and the Layer8 should correspond to the maximum non-zero value.
Data: The dput of data can be found on https://dl.dropboxusercontent.com/u/9267938/rcode.R
> head(data.frame(mytest))
datetime Year Month Day Hour Minute Second X1.07m X1.32m X1.57m X1.82m X2.07m X2.32m X2.57m X2.82m X3.07m
1 2014-08-03 12:40:00 2014 8 3 12 40 0 -0.079553637 -0.018856349 -0.022559778 -0.0278269427 -0.019816260 -0.01304108 -0.003394041 -0.010720688 NA
2 2014-08-03 12:50:00 2014 8 3 12 50 0 -0.001409806 0.006434559 0.013885671 0.0033940409 0.009665614 0.01176982 0.011130125 0.019991707 0.02997477
3 2014-08-03 13:00:00 2014 8 3 13 0 0 -0.006942835 -0.011130125 0.010715907 -0.0058745801 -0.005716650 0.01534520 0.030355206 0.024851408 0.04862646
4 2014-08-03 13:10:00 2014 8 3 13 10 0 -0.020586547 0.002935416 -0.016304143 -0.0001326389 -0.003896694 0.00361282 0.004723244 0.013947785 0.03787721
5 2014-08-03 13:20:00 2014 8 3 13 20 0 -0.028394300 -0.023132719 -0.001721911 -0.0139650391 -0.038460075 0.01749898 0.008466864 0.003630492 0.01442467
6 2014-08-03 13:30:00 2014 8 3 13 30 0 -0.034646511 -0.006791177 0.004064423 -0.0038792422 -0.015942808 -0.02029747 -0.014287663 0.007956902 0.01786172
X3.32m X3.57m X3.82m X4.07m X4.32m X4.57m X4.82m X5.07m X5.32m X5.57m X5.82m X6.07m X6.32m X6.57m X6.82m
1 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
2 0.05094966 0.04699597 0.032100892 0.02650842 0.045689389 0.0169759192 -0.006879327 -0.0187681077 -0.030404344 -0.04405705 -0.04501967 NA NA NA NA
3 0.04500833 0.01713256 0.006450535 0.02870071 0.019079580 0.0009741734 -0.024666588 -0.0409943643 -0.030201313 -0.03873463 -0.02893064 NA NA NA NA
4 0.03971244 0.05723497 0.039496306 0.03799276 0.012742073 0.0024111385 -0.023706420 -0.0188563490 -0.033791404 -0.04162619 -0.02979164 -0.045051204 NA NA NA
5 0.03269076 0.05125416 0.054766084 0.03625076 0.005988487 0.0020217180 -0.007510352 -0.0069913419 -0.006656083 -0.01630414 -0.01403812 -0.001580609 NA NA NA
6 0.01913708 0.03932811 0.048955209 0.04764632 0.037480601 0.0205218532 0.004171715 0.0009371753 -0.002468609 -0.04511612 -0.01263816 0.035861544 NA NA NA
X7.07m X7.32m X7.57m X7.82m X8.07m X8.32m X8.57m X8.82m X9.07m X9.32m X9.57m X9.82m X10.07m X10.32m X10.57m X10.82m X11.07m X11.32m X11.57m X11.82m
1 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
2 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
3 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
4 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
5 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
6 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
Sometimes the data points will be 20, 22, 25 points so the function should be written such that it would try to account those information and divide into 8 data values for each rows.
Rcode.R linked to dropbox has the code that has dput of mytest. It was pretty big to be posted here. So I posted a external link.
Info added
Each row would have different number of data. The motive is to convert them into 8 columns of data using averaging or linear interpolation.

Taking the question as a desire to collapse the values to means of eight equally spaced depths, dplyr and tidyr take us where we need to go:
library(dplyr)
library(tidyr)
mytest %>%
# melt to long form
gather(depth, value, -datetime:-Second, na.rm = TRUE) %>%
# clean depth to number
mutate(depth = extract_numeric(depth)) %>%
# group so cut levels are for each datetime
group_by(datetime) %>%
# group to keep columns; cut depth into 8 levels per group
group_by(datetime, levels = cut(depth, 8, paste0('level', 1:8))) %>%
# collapse groups by taking the mean
summarise(value = mean(value)) %>%
# re-spread new levels to wide form
spread(levels, value) %>%
# re-add other time columns dropped by summarise
inner_join(mytest %>% select(datetime:Second), .)
# Source: local data frame [20 x 15]
#
# datetime Year Month Day Hour Minute Second level1 level2
# (time) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl)
# 1 2014-08-03 12:40:00 2014 8 3 12 40 0 -0.079553637 -0.0188563490
# 2 2014-08-03 12:50:00 2014 8 3 12 50 0 0.006303474 0.0065298277
# 3 2014-08-03 13:00:00 2014 8 3 13 0 0 -0.002452351 -0.0057956151
# 4 2014-08-03 13:10:00 2014 8 3 13 10 0 -0.011318424 -0.0001388374
# 5 2014-08-03 13:20:00 2014 8 3 13 20 0 -0.017749644 -0.0116420430
# 6 2014-08-03 13:30:00 2014 8 3 13 30 0 -0.012457755 -0.0133731725
# 7 2014-08-03 13:40:00 2014 8 3 13 40 0 -0.020440875 -0.0253538846
# 8 2014-08-03 13:50:00 2014 8 3 13 50 0 -0.058681338 -0.0177194127
# 9 2014-08-03 14:00:00 2014 8 3 14 0 0 -0.037929680 -0.0211918383
# 10 2014-08-03 14:10:00 2014 8 3 14 10 0 -0.027045726 -0.0147261076
# 11 2014-08-03 14:20:00 2014 8 3 14 20 0 -0.048997399 -0.0290804019
# 12 2014-08-03 14:30:00 2014 8 3 14 30 0 -0.059110466 -0.0370898043
# 13 2014-08-03 14:40:00 2014 8 3 14 40 0 -0.067156867 -0.0138750287
# 14 2014-08-03 14:50:00 2014 8 3 14 50 0 -0.049762164 -0.0280648246
# 15 2014-08-03 15:00:00 2014 8 3 15 0 0 -0.028033559 -0.0245379952
# 16 2014-08-03 15:10:00 2014 8 3 15 10 0 -0.044087211 -0.0107995239
# 17 2014-08-03 15:20:00 2014 8 3 15 20 0 -0.028761973 -0.0113161242
# 18 2014-08-03 15:30:00 2014 8 3 15 30 0 -0.013476051 -0.0142316424
# 19 2014-08-03 15:40:00 2014 8 3 15 40 0 -0.012799297 -0.0135366710
# 20 2014-08-03 15:50:00 2014 8 3 15 50 0 -0.012238548 -0.0180806876
# Variables not shown: level3 (dbl), level4 (dbl), level5 (dbl), level6 (dbl), level7 (dbl),
# level8 (dbl)
Note that you should check that these data make sense in context; you've lost your depth data by scaling them.

Daily averages of all data frame variables including NA values with aggregate function

I want to calculate daily means of all variables in my dataframe which includes NA values. All my databases have a value every 30min, so I´m very interested in using the timestamp with aggregate function to obtain daily, weekly, monthly... aggregated data.
My dataframe is 37795 rows x 54 variables. I´ve tried two ways to do that, first option does not give me daily means cause I obtained too high values (not logical). Second option gives me almost all NA values. I do not what to do.
I write my dataframe head and code below.
head(data)
timestamp day month year.x hour minute doy.x rn_1_1_1 ppfd_1_1_1
1 2013-07-06 00:00:00 6 7 2013 0 0 187.000 -84.37381 0.754
2 2013-07-06 00:30:00 6 7 2013 0 30 187.020 -84.07990 0.808
3 2013-07-06 01:00:00 6 7 2013 1 0 187.041 -82.19991 0.808
4 2013-07-06 01:30:00 6 7 2013 1 30 187.062 -81.12341 0.831
5 2013-07-06 02:00:00 6 7 2013 2 0 187.083 -79.57474 0.708
6 2013-07-06 02:30:00 6 7 2013 2 30 187.104 -77.72460 0.639
ppfdr_1_1_1 p_rain_1_1_1 swc_1_1_1 swc_2_1_1 swc_3_1_1 air_pressure air_pressure.1
1 0.624 0 0.07230304 0.09577876 0.134602791 101212.4165 1012.124165
2 0.587 0 0.07233134 0.09569421 0.134479816 101181.8094 1011.818094
3 0.713 0 0.07242914 0.09566160 0.134203719 101166.0948 1011.660948
4 0.72 0 0.07252077 0.09563419 0.134149141 101144.6151 1011.446151
5 0.564 0 0.07261925 0.09560297 0.134095791 101144.8662 1011.448662
6 0.706 0 0.07271843 0.09557789 0.134037119 101144.5084 1011.445084
u_rot v_rot w_rot wind_speed u. h_scr_qc01_man
1 5.546047919 1.42E-14 4.76E-16 5.546047919 0.426515403 -28.07603618
2 5.122724997 6.94E-15 -8.00E-16 5.122724997 0.408213459 -34.39110979
3 5.248639421 4.56E-15 7.28E-17 5.248639421 0.393959075 -33.29033501
4 4.845257286 2.81E-14 -1.33E-17 4.845257286 0.365475898 -32.62427147
5 4.486426895 1.39E-14 -4.43E-16 4.486426895 0.335905384 -33.80219189
6 4.109603841 7.08E-15 -9.76E-16 4.109603841 0.312610588 -35.77289349
fco2_scr_qc01_man le_scr_qc01_man fco2_scr_qc0 fco2_scr_qc0_man date year.y time
1 -0.306504951 NA NA NA 06-jul-13 2013 0:00
2 -0.206266524 NA -0.206266524 -0.206266524 06-jul-13 2013 0:30
3 -0.268508139 NA -0.268508139 -0.268508139 06-jul-13 2013 1:00
4 -0.203804516 0.426531598 -0.203804516 -0.203804516 06-jul-13 2013 1:30
5 -0.217438742 -0.358248118 -0.217438742 -0.217438742 06-jul-13 2013 2:00
6 -0.193778528 2.571063044 -0.193778528 -0.193778528 06-jul-13 2013 2:30
doy_ent doy.y doy_cum doy_cum_ent mes nrecord bat panel_temp vwc_0.1
1 187 187.0000 187.0000 187 7 24 12.57 22.93 0.06284828
2 187 187.0208 187.0208 187 7 25 12.56 22.85 0.06267169
3 187 187.0417 187.0417 187 7 26 12.55 22.58 0.06261738
4 187 187.0625 187.0625 187 7 27 12.54 22.3 0.06247716
5 187 187.0833 187.0833 187 7 28 12.53 22.01 0.06249525
6 187 187.1042 187.1042 187 7 29 12.52 21.82 0.06236862
vwc_0.5 vwc_1.5 temp_0.1 temp_0.5 temp_1.5 tempsd_0.1 tempsd_0.5 tempsd_1.5
1 0.07569027 0.1007845 30.9 28.96 25.14 0.372 0.961 0.767
2 0.07569027 0.1007743 30.8 28.85 24.99 0.181 1.361 1.087
3 0.07568554 0.1008558 30.53 28.8 25.03 0.98 1.476 0.351
4 0.07559577 0.1008507 30.52 29.09 25.11 0.186 0.229 0.556
5 0.07559577 0.1007743 30.11 29.09 24.87 1.331 0.191 0.954
6 0.07556271 0.1007285 30.15 29.33 25.04 1.447 1.078 0.2
pair pair_avg CO2_0.1 CO2_0.5 CO2_1.5 DCO2_0.1 DCO2_0.5
1 101.2124 101.2118 1161.592832 3275.1134 4888.231603 -24.67422109 34.88538221
2 101.1818 101.2131 1168.144925 3338.24016 4941.418642 6.55209301 63.12675931
3 101.1661 101.2090 1201.049131 3435.235974 5012.525851 32.90420541 96.9958144
4 101.1446 101.2007 1268.613941 3556.723878 5092.96558 67.56481067 121.4879035
5 101.1449 101.1906 1364.315214 3680.188043 5164.795759 95.7012722 123.464165
6 101.1445 101.1805 1472.975286 3808.988677 5236.40855 108.6600723 128.8006346
DCO2_1.5
1 31.30293041
2 53.18703947
3 71.10720845
4 80.43972916
5 71.83017884
6 71.61279156
## Daily avg - OPTION 1
data$timestamp <- as.POSIXct(data$timestamp, format = "%d/%m/%Y %H:%M",tz ="GMT")
> dates <- format(data$timestamp,"%Y/%m/%d",tz = "GMT")
> datadates <- cbind(data,dates)
> dailydata_avg <- aggregate(. ~ dates, datadates, FUN=mean, na.rm=TRUE, na.action = "na.pass")
head(dailydata_avg)
dates timestamp day month year.x hour minute doy.x rn_1_1_1 ppfd_1_1_1
1 2013/07/06 1373111100 6 7 2013 11.5 15 187.489 159.7788 3580.562
2 2013/07/07 1373197500 7 7 2013 11.5 15 188.489 154.0925 3506.688
3 2013/07/08 1373283900 8 7 2013 11.5 15 189.489 152.5259 3460.667
4 2013/07/09 1373370300 9 7 2013 11.5 15 190.489 131.1619 2965.250
5 2013/07/10 1373456700 10 7 2013 11.5 15 191.489 136.7853 3171.958
6 2013/07/11 1373543100 11 7 2013 11.5 15 192.489 145.2757 3282.167
ppfdr_1_1_1 p_rain_1_1_1 swc_1_1_1 swc_2_1_1 swc_3_1_1 air_pressure air_pressure.1
1 2552.396 1.0000 0.07095847 0.09606378 18341.81 25940.167 25940.167
2 2532.542 1.0000 0.06994341 0.09502167 18065.98 24891.000 24891.000
3 2523.562 1.0000 0.06860553 0.09379282 17777.02 23107.271 23107.271
4 2336.000 1.0000 0.06717054 0.09268716 17526.50 19309.500 19309.500
5 2607.229 1.0625 0.06620048 0.09166904 17275.56 8385.646 8385.646
6 2484.521 1.0000 0.06562964 0.09083684 17028.94 3535.438 3535.438
u_rot v_rot w_rot wind_speed u. h_scr_qc01_man fco2_scr_qc01_man
1 32167.83 2215.875 2041.354 32167.83 28531.44 18197.75 15365.65
2 30878.27 1911.312 1939.917 30878.27 26929.62 17605.52 14955.56
3 26052.96 2261.417 2116.458 26052.96 23305.83 19167.98 18399.33
4 17284.04 1987.438 2139.083 17284.04 17704.35 20349.92 18137.65
5 12028.06 2053.812 1960.417 12028.06 15670.00 21997.83 21120.19
6 15607.50 1997.417 1907.646 15607.50 15384.56 18000.94 18810.62
le_scr_qc01_man fco2_scr_qc0 fco2_scr_qc0_man date year.y time doy_ent doy.y
1 17409.67 13032.10 13027.90 137 2013 44.5 187 187.4896
2 15524.38 12077.17 12072.92 163 2013 44.5 188 188.4896
3 16407.71 14775.94 14770.56 189 2013 44.5 189 189.4896
4 16788.04 15024.79 15019.02 215 2013 44.5 190 190.4896
5 17955.58 17737.25 17730.75 241 2013 44.5 191 191.4896
6 14610.02 16605.48 16599.33 267 2013 44.5 192 192.4896
doy_cum doy_cum_ent mes nrecord bat panel_temp vwc_0.1 vwc_0.5 vwc_1.5
1 187.4896 187.5 7 28966.375 111.5208 1836.250 4638.833 4594.396 37.35417
2 188.4896 188.5 7 20801.417 111.7292 1900.812 4656.875 4392.979 26.68750
3 189.4896 189.5 7 4394.500 110.6042 1934.792 4675.604 4238.229 65.20833
4 190.4896 190.5 7 9467.708 104.0000 2090.896 4776.521 4178.729 54.12500
5 191.4896 191.5 7 14796.375 109.7500 2145.875 4907.292 4161.312 108.39583
6 192.4896 192.5 7 20127.958 109.3125 1934.375 4876.021 4123.458 143.10417
temp_0.1 temp_0.5 temp_1.5 tempsd_0.1 tempsd_0.5 tempsd_1.5 pair pair_avg CO2_0.1
1 2018.438 1565.812 797.8750 470.8125 474.3958 508.8333 101.1268 101.1323 10400.27
2 1998.438 1574.000 783.1875 478.3333 460.4583 566.0208 101.0764 101.0789 11292.75
3 1994.833 1568.104 780.2083 463.8125 453.1667 488.5625 100.9967 101.0036 13288.25
4 2042.625 1564.875 780.1667 465.0000 599.2708 437.6042 100.8520 100.8665 16156.60
5 2114.708 1576.729 780.5000 471.5833 406.5417 484.6875 100.4828 100.5169 18656.50
6 2124.604 1591.125 781.8125 516.7500 530.3333 510.7500 100.3025 100.2947 14586.60
CO2_0.5 CO2_1.5 DCO2_0.1 DCO2_0.5 DCO2_1.5
1 26360.38 34371.31 19795.81 20637.94 27123.92
2 26939.60 34558.17 18838.38 20464.56 20452.58
3 27603.06 34608.31 17413.15 19998.02 22754.85
4 28572.69 34678.38 19294.62 21894.92 18379.62
5 28983.29 34644.15 20251.17 20409.58 22077.40
6 28236.12 34736.67 17031.02 18852.04 19684.69`
## Daily avg - OPTION 2
data$timestamp <- as.POSIXct(data$timestamp, format = "%d/%m/%Y %H:%M",tz ="GMT")
datatime <- data$timestamp
dailydata_avg <- aggregate( data,
by = list('DATES'= format(datatime,'%Y%m%d' )),
FUN = mean, na.rm=T)
I obtain this console message:
1: In mean.default(X[[i]], ...) :
argument is not numeric or logical: returning NA
2: In mean.default(X[[i]], ...) :
argument is not numeric or logical: returning NA
3: In mean.default(X[[i]], ...) :
argument is not numeric or logical: returning NA
head(dailydata_avg)
DATES timestamp day month year.x hour minute doy.x rn_1_1_1 ppfd_1_1_1
1 20130706 2013-07-06 13:45:00 6 7 2013 11.5 15 187.489 159.7788 NA
2 20130707 2013-07-07 13:45:00 7 7 2013 11.5 15 188.489 154.0925 NA
3 20130708 2013-07-08 13:45:00 8 7 2013 11.5 15 189.489 152.5259 NA
4 20130709 2013-07-09 13:45:00 9 7 2013 11.5 15 190.489 131.1619 NA
5 20130710 2013-07-10 13:45:00 10 7 2013 11.5 15 191.489 136.7853 NA
6 20130711 2013-07-11 13:45:00 11 7 2013 11.5 15 192.489 145.2757 NA
ppfdr_1_1_1 p_rain_1_1_1 swc_1_1_1 swc_2_1_1 swc_3_1_1 air_pressure air_pressure.1
1 NA NA 0.07095847 0.09606378 NA NA NA
2 NA NA 0.06994341 0.09502167 NA NA NA
3 NA NA 0.06860553 0.09379282 NA NA NA
4 NA NA 0.06717054 0.09268716 NA NA NA
5 NA NA 0.06620048 0.09166904 NA NA NA
6 NA NA 0.06562964 0.09083684 NA NA NA
u_rot v_rot w_rot wind_speed u. h_scr_qc01_man fco2_scr_qc01_man le_scr_qc01_man
1 NA NA NA NA NA NA NA NA
2 NA NA NA NA NA NA NA NA
3 NA NA NA NA NA NA NA NA
4 NA NA NA NA NA NA NA NA
5 NA NA NA NA NA NA NA NA
6 NA NA NA NA NA NA NA NA
fco2_scr_qc0 fco2_scr_qc0_man date year.y time doy_ent doy.y doy_cum doy_cum_ent
1 NA NA NA 2013 NA 187 187.4896 187.4896 187.5
2 NA NA NA 2013 NA 188 188.4896 188.4896 188.5
3 NA NA NA 2013 NA 189 189.4896 189.4896 189.5
4 NA NA NA 2013 NA 190 190.4896 190.4896 190.5
5 NA NA NA 2013 NA 191 191.4896 191.4896 191.5
6 NA NA NA 2013 NA 192 192.4896 192.4896 192.5
mes nrecord bat panel_temp vwc_0.1 vwc_0.5 vwc_1.5 temp_0.1 temp_0.5 temp_1.5
1 7 NA NA NA NA NA NA NA NA NA
2 7 NA NA NA NA NA NA NA NA NA
3 7 NA NA NA NA NA NA NA NA NA
4 7 NA NA NA NA NA NA NA NA NA
5 7 NA NA NA NA NA NA NA NA NA
6 7 NA NA NA NA NA NA NA NA NA
tempsd_0.1 tempsd_0.5 tempsd_1.5 pair pair_avg CO2_0.1 CO2_0.5 CO2_1.5 DCO2_0.1
1 NA NA NA 101.1268 101.1323 NA NA NA NA
2 NA NA NA 101.0764 101.0789 NA NA NA NA
3 NA NA NA 100.9967 101.0036 NA NA NA NA
4 NA NA NA 100.8520 100.8665 NA NA NA NA
5 NA NA NA 100.4828 100.5169 NA NA NA NA
6 NA NA NA 100.3025 100.2947 NA NA NA NA
DCO2_0.5 DCO2_1.5
1 NA NA
2 NA NA
3 NA NA
4 NA NA
5 NA NA
6 NA NA
How could I do it?
Thanks!!

I didn't use the aggregate function, I used the tapply one.
This is the code, that deals with NA's, I came up with:
# create a sequence of DateTime with half-hourly data
DateTime <- seq.POSIXt(from = as.POSIXct("2015-05-01 00:00:00", tz = "Etc/GMT+12"),
to = as.POSIXct("2015-05-30 23:59:00", tz = "Etc/GMT+12"), by = 1800)
# create some dummy data of the same length as DateTime vector
aa <- runif(1440, 5.0, 7.5)
bb <- NA
df <- data.frame(DateTime, aa, bb)
# replace a cell with NA in the "a" column
df[19,2] <- NA # dataframe = df, row = 19, column = 2
# create DateHour column to use later
df$DateHour <- paste(format(df$DateTime, "%Y/%m/%d"), format(df$DateTime, "%H"), sep = " ")
View(df)
# Hourly means
# Calculate hourly mean values
aa.HourlyMean <- tapply(df$aa, df$DateHour, mean, na.rm = TRUE)
# convert the vector to dataframe
aa.HourlyMean <- data.frame(aa.HourlyMean)
# Extract the DateHour column from the "aa" dataframe
aa.HourlyMean$DateHour <- row.names(aa.HourlyMean);
# Delete rownames of "aa" dataframe
row.names(aa.HourlyMean) <- NULL
# Create a tidy DateTime column
aa.HourlyMean$DateTime <- as.POSIXct(aa.HourlyMean$DateHour, "%Y/%m/%d %H", tz = "Etc/GMT+12")
# change to a tidy dataframe
aa.HourlyMean <- aa.HourlyMean[,c(3,2,1)]
# You can delete any column (for example, DateHour) by
# aa.HourlyMean$Date <- NULL
# You can rename a column with "plyr" package by
# rename(aa.HourlyMean)[3] <- "NewColumnName"
# View the hourly mean of the "aa" dataframe
View(aa.HourlyMean)
# You can do the same with the "bb" vector
bb.HourlyMean <- tapply(df$bb, df$DateHour, mean, na.rm = TRUE)
bb.HourlyMean <- data.frame(bb.HourlyMean)
# View the hourly mean of the "bb" vector
View(bb.HourlyMean)
# /Hourly means
You then can combine in one dataframe the aa.HourlyMean and bb.HourlyMean vectors.
# Daily means
df$Date <- format(df$DateTime, "%Y/%m/%d")
aa.DailyMean <- tapply(df$aa, df$Days, mean, na.rm = TRUE)
aa.DailyMean <- data.frame(aa.DailyMean)
aa.DailyMean$Date <- row.names(aa.DailyMean); row.names(aa.DailyMean) <- NULL
aa.DailyMean <- aa.DailyMean[,c(2,1)]
View(aa.DailyMean)
# /Daily means
# Weekly means
df$YearWeek <- paste(format(df$DateTime, "%Y"), strftime(DateTime, format = "%W"), sep = " ")
aa.WeeklyMean <- tapply(df$aa, df$YearWeek, mean, na.rm = TRUE)
aa.WeeklyMean <- data.frame(aa.WeeklyMean)
aa.WeeklyMean$YearWeek <- row.names(aa.WeeklyMean); row.names(aa.WeeklyMean) <- NULL
aa.WeeklyMean <- aa.WeeklyMean[,c(2,1)]
View(aa.WeeklyMean)
# /Weekly means
I created the mean values for hourly, daily and weekly observations but you get the idea how to create the monthly, yearly, ... ones.

R Fill cells with previous data

I have a table like the following:
days Debit loaddate
1 23/01/2014 138470289.4 23/01/2014
2 24/01/2014 NA NA
3 25/01/2014 NA NA
4 26/01/2014 NA NA
5 27/01/2014 NA NA
one row for each day and then in the columns loaddate after a few NA another date appears:
28 19/02/2014 NA NA
29 20/02/2014 NA NA
30 21/02/2014 NA NA
31 22/02/2014 9090967.9 22/02/2014
32 23/02/2014 NA NA
33 24/02/2014 308083.5 24/02/2014
I would like to replace each NA in loaddate column with the previous date in loaddate.
I tried:
for(i in 1:nrow(data3))
{ if (!is.na(data3[i,'Debit']))
{data3[i,'loaddate1']<-as.Date(data3[i,'loaddate'], format='%Y-%m-%d')}
else {data3[i,'loaddate1']<-data3[i-1,'loaddate1']}
}
But I got the wrong format:
> head(data3)
days Debit loaddate loaddate1
1 2014-01-23 138470289 2014-01-23 16093
2 2014-01-24 NA <NA> 16093
3 2014-01-25 NA <NA> 16093
4 2014-01-26 NA <NA> 16093
5 2014-01-27 NA <NA> 16093
6 2014-01-28 NA <NA> 16093
I need to get the date format also. If I do:
for(i in 1:nrow(data3))
{ if (!is.na(data3[i,'Debit']))
{data3[i,'loaddate1']<-as.Date(data3[i,'loaddate'], format='%Y-%m-%d')}
else {data3[i,'loaddate1']<-as.Date(data3[i-1,'loaddate1'], format='%Y-%m-%d')}
}
I got the wrong result (with NA).
days Debit loaddate loaddate1
1 2014-01-23 138470289 2014-01-23 16093
2 2014-01-24 NA <NA> <NA>
3 2014-01-25 NA <NA> <NA>
4 2014-01-26 NA <NA> <NA>
5 2014-01-27 NA <NA> <NA>
6 2014-01-28 NA <NA> <NA>
How can I get the right result and with the right format?
Also, Is there a better way to do this replacement? I mean without a loop.
Thanks.

Try zoo::na.locf and make sure to use the appropriate date format:
library(zoo)
data3$loaddate <- as.Date(na.locf(data3$loaddate), format='%d/%m/%Y'))

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