I want to subset my dataframe from Sept 2017 to April 2018. My dataframe is like this:-
Year Month Day Avg_Temp
2017 8 31 20
2017 9 1 22
.
.
.
2018 4 30 26
2018 5 1 30
I want that my dataset from 1 Sept 2017 to 30 April 2018.
Year Month Day Avg_Temp
2017 9 1 22
.
.
.
2018 4 30 26
based on just the year I am to do subset.
df <-df[df$YEAR>="2017" & df$YEAR<="2018", ]
But I need to subset from month as well. Any help would be great
Try this option:
df <- df[(df$Year == 2017 & df$Month >= 9) |
(df$Year == 2018 & df$Month <= 4), ]
By the way, you might want to consider storing your dates as a proper date type, including a day component.
Here is a dplyr approach:
require(tidyverse)
df<-data.frame(Year=c(2018,2017,2017,2017,2018,2018,2018),
Month=c(9,8,10,4,9,3,4),Day=c(13,12,14,15,17,15,14))
df %>%
filter(Year==2017&Month>=9|Year==2018&Month<=4)
Which Yields this:
Year Month Day
1 2017 9 14
2 2018 3 15
3 2018 4 14
Perhaps it would be easier if the three date components were encoded in one Date column :
df$Date <- as.Date(paste(df$Year, df$Month, df$Date, sep = '-'))
df$Year <- NULL
df$Month <- NULL
df <- df[df$Date > as.Date('2017-09-01') & df$Date < as.Date('2018-04-01'), ]
Related
This is my data:
Year1 <- c(2015,2013,2012,2018)
Year2 <- c(2017,2015,2014,2020)
my_data <- data.frame(Year1, Year2)
I need an if statement that returns 1 when year 1 equals 2015 OR 2016 AND year 2 is greater than 2016. Currently, my code looks like this:
my_data <- my_data %>%
mutate(Y_2016=ifelse(my_data$Year1==2015|2016 & my_data$Year1>2016,1,0))
But this does not work and only seems to check the condition if Year 2 is greater than 2016, since it returns 1 even for the last row when Year 1 is 2018 and Year 2 is 2020.
Thank you for your help!
Instead of my_data$Year1==2015|2016, use %in% like my_data$Year1 %in% c(2015,2016).
Typo in my_data$Year1>2016
As you using dplyr you do not need to specify every variable with $ like my_data$...
my_data%>%
mutate(Y_2016=ifelse(Year1 %in% c(2015,2016) & Year2>2016,1,0))
Year1 Year2 Y_2016
1 2015 2017 1
2 2013 2015 0
3 2012 2014 0
4 2018 2020 0
I so have the following data frame
customerid
payment_month
payment_date
bill_month
charges
1
January
22
January
30
1
February
15
February
21
1
March
2
March
33
1
May
4
April
43
1
May
4
May
23
1
June
13
June
32
2
January
12
January
45
2
February
15
February
56
2
March
2
March
67
2
April
4
April
65
2
May
4
May
54
2
June
13
June
68
3
January
25
January
45
3
February
26
February
56
3
March
30
March
67
3
April
1
April
65
3
June
1
May
54
3
June
1
June
68
(the id data is much larger) I want to calculate payment efficiency using the following function,
efficiency = (amount paid not late / total bill amount)*100
not late is paying no later than the 21st day of the bill's month. (paying January's bill on the 22nd of January is considered as late)
I want to calculate the efficiency of each customer with the expected output of
customerid
effectivity
1
59.90
2
100
3
37.46
I have tried using the following code to calculate for one id and it works. but I want to apply and assign it to the entire group id and summarize it into 1 column (effectivity) and 1 row per ID. I have tried using group by, aggregate and ifelse functions but nothing works. What should I do?
df1 <- filter(df, (payment_month!=bill_month & id==1) | (payment_month==bill_month & payment_date > 21 & id==1) )
df2 <-filter(df, id==1001)
x <- sum(df1$charges)
x <- sum(df2$charges)
100-(x/y)*100
An option using dplyr
library(dplyr)
df %>%
group_by(customerid) %>%
summarise(
effectivity = sum(
charges[payment_date <= 21 & payment_month == bill_month]) / sum(charges) * 100,
.groups = "drop")
## A tibble: 3 x 2
#customerid effectivity
# <int> <dbl>
#1 1 59.9
#2 2 100
#3 3 37.5
df %>%
group_by(customerid) %>%
mutate(totalperid = sum(charges)) %>%
mutate(pay_month_number = match(payment_month , month.name),
bill_month_number = match(bill_month , month.name)) %>%
mutate(nolate = ifelse(pay_month_number > bill_month_number, TRUE, FALSE)) %>%
summarise(efficiency = case_when(nolate = TRUE ~ (charges/totalperid)*100))
I have a data set with the variable 'months' from 1 to 12, but need to change them to the month names. i.e "1" needs to be January and so on. Whats the easiest way to do this?
R has an inbuilt vector called month.name for your purpose you could do something like the following:
# Some dummy data
set.seed(1)
df <- data.frame(
month = sample(1:12, size = 10)
)
# Now use your integer month to subset month.name
df$month2 <- month.name[df$month] # Also has month.abb
df
month month2
1 9 September
2 4 April
3 7 July
4 1 January
5 2 February
6 5 May
7 3 March
8 8 August
9 6 June
10 11 November
I have a dataframe like this. The time span is 10 years. Because it's Chinese market data, and China has Lunar Holidays. So each year have different holiday times in terms of the western calendar.
When it is a holiday, the stock market does not open, so it is a non-trading day. Weekends are non-trading days too.
I want to find out which month of which year has the least number of trading days, and most importantly, what number is that.
There are not repeated days.
date change open high low close volume
1 1995-01-03 -1.233 637.72 647.71 630.53 639.88 234518
2 1995-01-04 2.177 641.90 655.51 638.86 653.81 422220
3 1995-01-05 -1.058 656.20 657.45 645.81 646.89 430123
4 1995-01-06 -0.948 642.75 643.89 636.33 640.76 487482
5 1995-01-09 -2.308 637.52 637.55 625.04 625.97 509851
6 1995-01-10 -2.503 616.16 617.60 607.06 610.30 606925
If there are not repeated days, you can count days per month and year by:
library(data.table) "maxx"))), .Names = c("X2005", "X2006", "X2007", "X2008"))
library(lubridate)
dt <- as.data.table(dt)
dt_days <- dt[, .(count_day=.N), by=.(year(date), month(date))]
Then you only need to do this to get the min:
dt_days[count_day==min(count_day)]
The chron and bizdays packages deal with business days but neither actually contains a usable calendar of holidays limiting their usefulness.
We will use chron below assuming you have defined the .Holidays vector of dates that are holidays. (If you run the code below without doing that only weekdays will be regarded as business days as the default .Holidays vector supplied by chron has very few dates in it.) DF has 120 rows (one row for each year/month) and the last line subsets that to just the month in each year having least business days.
library(chron)
library(zoo)
st <- as.yearmon("2001-01")
en <- as.yearmon("2010-12")
ym <- seq(st, en, 1/12) # sequence of year/months of interest
# no of business days in each yearmonth
busdays <- sapply(ym, function(x) {
s <- seq(as.Date(x), as.Date(x, frac = 1), "day")
sum(!is.weekend(s) & !is.holiday(s))
})
# data frame with one row per year/month
yr <- as.integer(ym)
DF <- data.frame(year = yr, month = cycle(ym), yearmon = ym, busdays)
# data frame with one row per year
wx.min <- ave(busdays, yr, FUN = function(x) which.min(x) == seq_along(x))
DF[wx.min == 1, ]
giving:
year month yearmon busdays
2 2001 2 Feb 2001 20
14 2002 2 Feb 2002 20
26 2003 2 Feb 2003 20
38 2004 2 Feb 2004 20
50 2005 2 Feb 2005 20
62 2006 2 Feb 2006 20
74 2007 2 Feb 2007 20
95 2008 11 Nov 2008 20
98 2009 2 Feb 2009 20
110 2010 2 Feb 2010 20
I want to return the final row for each subsection of a dataframe. I'm aware of the ddply and aggregate functions, but they are not giving the expected output in this case, as the column by which I split the data has recurring names.
For example, in df:
year <- rep(c(2011, 2012, 2013), each=12)
season <- rep(c("Spring", "Summer", "Autumn", "Winter"), each=3)
allseason <- rep(season, 3)
temp <- rnorm(36, mean = 61, sd = 10)
df <- data.frame(year, allseason, temp)
I want to return the final temp reading at the end of every season. When I run either
final1 <- aggregate(df, list(df$allseason), tail, 1)
or
final2 <- ddply(df, .(allseason), tail, 1)
I get only the final 4 seasons (i.e. those of 2013). The function seems to stop there and does not go back to previous years/seasons. My intended output is a data frame with 12 rows * 3 columns.
All help appreciated!
*I notice that in the df created here, the allseasons column is designated as a factor with 4 levels, whereas this is not the case in my original dataframe.
In your ddply code, you only forgot to also group by year:
With plyr:
library(plyr)
ddply(df, .(year, allseason), tail, 1)
Or with dplyr
library(dplyr)
df %>%
group_by(year, allseason) %>%
do(tail(.,1))
Or if you want a base R alternative you can use ave:
df[with(df, ave(year, list(year, allseason), FUN = seq_along)) == 3,]
Result:
# year allseason temp
#1 2011 Autumn 63.40626
#2 2011 Spring 59.69441
#3 2011 Summer 42.33252
#4 2011 Winter 79.10926
#5 2012 Autumn 63.14974
#6 2012 Spring 60.32811
#7 2012 Summer 67.57364
#8 2012 Winter 61.39100
#9 2013 Autumn 50.30501
#10 2013 Spring 61.43044
#11 2013 Summer 55.16605
#12 2013 Winter 69.37070
Note that the output will contain the same rows in each case, only the ordering may differ.
And just to add to #beginneR's answer, your aggregate solution should look like:
aggregate(temp ~ allseason + year, data = df, tail, 1)
# or:
with(df, aggregate(temp, list(allseason, year), tail, 1))
Result:
allseason year temp
1 Autumn 2011 64.51539
2 Spring 2011 45.14341
3 Summer 2011 62.29240
4 Winter 2011 47.97461
5 Autumn 2012 43.16781
6 Spring 2012 80.02419
7 Summer 2012 72.31149
8 Winter 2012 45.58344
9 Autumn 2013 55.92607
10 Spring 2013 52.06778
11 Summer 2013 51.01308
12 Winter 2013 53.22452