When Time Series data is offset and incomplete
A new client at work has a fiscal calendar that starts in March and ends in February of the following year:
fiscalMonthLabels <- c("March", "April", "May", "June",
"July", "August", "September", "October",
"November", "December", "January", "February")
But, because they are new, we only have a few months' worth of data:
library(lubridate)
rawDate <- c("2018-09-01", "2018-10-01", "2018-11-01")
actualMonth <- month(rawDate)
newMonth <- rep(0, length(actualMonth))
for (i in 1:length(actualMonth)) {
if (actualMonth[i] == 1) {newMonth[i] <- 11}
else if (actualMonth[i] == 2) {newMonth[i] <- 12}
else {newMonth[i] <- actualMonth[i] - 2}
}
revenue <- c(123, 456, 789)
df <- data.frame(rawDate, actualMonth, newMonth, revenue)
df
rawDate actualMonth newMonth revenue
1 2018-09-01 9 7 123
2 2018-10-01 10 8 456
3 2018-11-01 11 9 789
So when I try to create a new factor with the fiscal month, this is the error I get:
fiscalMonth <- factor(newMonth, labels = fiscalMonthLabels)
Error in factor(newMonth, labels = fiscalMonthLabels) :
invalid 'labels'; length 12 should be 1 or 3
It seems like the factor command is looking for actualMonth to contain all twelve possible values in it. How do I get around this issue?
r
asked Nov 12 at 16:43
mmyoung77
356211
add a comment |
A new client at work has a fiscal calendar that starts in March and ends in February of the following year:
fiscalMonthLabels <- c("March", "April", "May", "June",
"July", "August", "September", "October",
"November", "December", "January", "February")
But, because they are new, we only have a few months' worth of data:
library(lubridate)
rawDate <- c("2018-09-01", "2018-10-01", "2018-11-01")
actualMonth <- month(rawDate)
newMonth <- rep(0, length(actualMonth))
for (i in 1:length(actualMonth)) {
if (actualMonth[i] == 1) {newMonth[i] <- 11}
else if (actualMonth[i] == 2) {newMonth[i] <- 12}
else {newMonth[i] <- actualMonth[i] - 2}
}
revenue <- c(123, 456, 789)
df <- data.frame(rawDate, actualMonth, newMonth, revenue)
df
rawDate actualMonth newMonth revenue
1 2018-09-01 9 7 123
2 2018-10-01 10 8 456
3 2018-11-01 11 9 789
So when I try to create a new factor with the fiscal month, this is the error I get:
fiscalMonth <- factor(newMonth, labels = fiscalMonthLabels)
Error in factor(newMonth, labels = fiscalMonthLabels) :
invalid 'labels'; length 12 should be 1 or 3
It seems like the factor command is looking for actualMonth to contain all twelve possible values in it. How do I get around this issue?
r
asked Nov 12 at 16:43
mmyoung77
356211
I made an edit, just so it's clear that the new months will be numbered appropriately to reflect the calendar shift.
– mmyoung77
Nov 12 at 16:57
add a comment |
A new client at work has a fiscal calendar that starts in March and ends in February of the following year:
fiscalMonthLabels <- c("March", "April", "May", "June",
"July", "August", "September", "October",
"November", "December", "January", "February")
But, because they are new, we only have a few months' worth of data:
library(lubridate)
rawDate <- c("2018-09-01", "2018-10-01", "2018-11-01")
actualMonth <- month(rawDate)
newMonth <- rep(0, length(actualMonth))
for (i in 1:length(actualMonth)) {
if (actualMonth[i] == 1) {newMonth[i] <- 11}
else if (actualMonth[i] == 2) {newMonth[i] <- 12}
else {newMonth[i] <- actualMonth[i] - 2}
}
revenue <- c(123, 456, 789)
df <- data.frame(rawDate, actualMonth, newMonth, revenue)
df
rawDate actualMonth newMonth revenue
1 2018-09-01 9 7 123
2 2018-10-01 10 8 456
3 2018-11-01 11 9 789
So when I try to create a new factor with the fiscal month, this is the error I get:
fiscalMonth <- factor(newMonth, labels = fiscalMonthLabels)
Error in factor(newMonth, labels = fiscalMonthLabels) :
invalid 'labels'; length 12 should be 1 or 3
It seems like the factor command is looking for actualMonth to contain all twelve possible values in it. How do I get around this issue?
r
asked Nov 12 at 16:43
mmyoung77
356211
A new client at work has a fiscal calendar that starts in March and ends in February of the following year:
fiscalMonthLabels <- c("March", "April", "May", "June",
"July", "August", "September", "October",
"November", "December", "January", "February")
But, because they are new, we only have a few months' worth of data:
library(lubridate)
rawDate <- c("2018-09-01", "2018-10-01", "2018-11-01")
actualMonth <- month(rawDate)
newMonth <- rep(0, length(actualMonth))
for (i in 1:length(actualMonth)) {
if (actualMonth[i] == 1) {newMonth[i] <- 11}
else if (actualMonth[i] == 2) {newMonth[i] <- 12}
else {newMonth[i] <- actualMonth[i] - 2}
}
revenue <- c(123, 456, 789)
df <- data.frame(rawDate, actualMonth, newMonth, revenue)
df
rawDate actualMonth newMonth revenue
1 2018-09-01 9 7 123
2 2018-10-01 10 8 456
3 2018-11-01 11 9 789
So when I try to create a new factor with the fiscal month, this is the error I get:
fiscalMonth <- factor(newMonth, labels = fiscalMonthLabels)
Error in factor(newMonth, labels = fiscalMonthLabels) :
invalid 'labels'; length 12 should be 1 or 3
It seems like the factor command is looking for actualMonth to contain all twelve possible values in it. How do I get around this issue?
r
r
asked Nov 12 at 16:43
mmyoung77
356211
asked Nov 12 at 16:43
mmyoung77
356211
edited Nov 12 at 16:56
asked Nov 12 at 16:43
mmyoung77
356211
asked Nov 12 at 16:43
mmyoung77
356211
asked Nov 12 at 16:43
mmyoung77
356211
356211
I made an edit, just so it's clear that the new months will be numbered appropriately to reflect the calendar shift.
– mmyoung77
Nov 12 at 16:57
add a comment |
I made an edit, just so it's clear that the new months will be numbered appropriately to reflect the calendar shift.
– mmyoung77
Nov 12 at 16:57
I made an edit, just so it's clear that the new months will be numbered appropriately to reflect the calendar shift.
– mmyoung77
Nov 12 at 16:57
I made an edit, just so it's clear that the new months will be numbered appropriately to reflect the calendar shift.
– mmyoung77
Nov 12 at 16:57
add a comment |
1 Answer
1
active
oldest
votes
You're going to want to assign the levels too:
fiscalMonth <- factor(actualMonth, levels = 1:12, labels = fiscalMonthLabels)
fiscalMonth
[1] November December January
Levels: March April May June July August September October November December January February
Alternatively, since you're using lubridate::month, you could just pass the label argument to month, which will return an ordered factor:
fiscalMonth <- month(actualMonth, label = TRUE)
[1] Sep Oct Nov
Levels: Jan < Feb < Mar < Apr < May < Jun < Jul < Aug < Sep < Oct < Nov < Dec
answered Nov 12 at 16:49
JasonAizkalns
10.9k42976
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
You're going to want to assign the levels too:
fiscalMonth <- factor(actualMonth, levels = 1:12, labels = fiscalMonthLabels)
fiscalMonth
[1] November December January
Levels: March April May June July August September October November December January February
Alternatively, since you're using lubridate::month, you could just pass the label argument to month, which will return an ordered factor:
fiscalMonth <- month(actualMonth, label = TRUE)
[1] Sep Oct Nov
Levels: Jan < Feb < Mar < Apr < May < Jun < Jul < Aug < Sep < Oct < Nov < Dec
answered Nov 12 at 16:49
JasonAizkalns
10.9k42976
add a comment |
You're going to want to assign the levels too:
fiscalMonth <- factor(actualMonth, levels = 1:12, labels = fiscalMonthLabels)
fiscalMonth
[1] November December January
Levels: March April May June July August September October November December January February
Alternatively, since you're using lubridate::month, you could just pass the label argument to month, which will return an ordered factor:
fiscalMonth <- month(actualMonth, label = TRUE)
[1] Sep Oct Nov
Levels: Jan < Feb < Mar < Apr < May < Jun < Jul < Aug < Sep < Oct < Nov < Dec
answered Nov 12 at 16:49
JasonAizkalns
10.9k42976
add a comment |
You're going to want to assign the levels too:
fiscalMonth <- factor(actualMonth, levels = 1:12, labels = fiscalMonthLabels)
fiscalMonth
[1] November December January
Levels: March April May June July August September October November December January February
Alternatively, since you're using lubridate::month, you could just pass the label argument to month, which will return an ordered factor:
fiscalMonth <- month(actualMonth, label = TRUE)
[1] Sep Oct Nov
Levels: Jan < Feb < Mar < Apr < May < Jun < Jul < Aug < Sep < Oct < Nov < Dec
answered Nov 12 at 16:49
JasonAizkalns
10.9k42976
You're going to want to assign the levels too:
fiscalMonth <- factor(actualMonth, levels = 1:12, labels = fiscalMonthLabels)
fiscalMonth
[1] November December January
Levels: March April May June July August September October November December January February
Alternatively, since you're using lubridate::month, you could just pass the label argument to month, which will return an ordered factor:
fiscalMonth <- month(actualMonth, label = TRUE)
[1] Sep Oct Nov
Levels: Jan < Feb < Mar < Apr < May < Jun < Jul < Aug < Sep < Oct < Nov < Dec
answered Nov 12 at 16:49
JasonAizkalns
10.9k42976
answered Nov 12 at 16:49
JasonAizkalns
10.9k42976
answered Nov 12 at 16:49
JasonAizkalns
10.9k42976
answered Nov 12 at 16:49
JasonAizkalns
10.9k42976
10.9k42976
add a comment |
add a comment |
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I made an edit, just so it's clear that the new months will be numbered appropriately to reflect the calendar shift.
– mmyoung77
Nov 12 at 16:57