I want to categorise entries in a dataframe, e.g. 1970, 1971, 1972 turn into “70s”
0
That's what I have:
numbers any_data year period
1 ab 1974 <NA
2 cd 1975 <NA
3 ef 1985 <NA
4 gh 1960 <NA
5 ij 1955 <NA
...and that is what I want:
numbers any_data year period
1 ab 1974 "70s"
2 cd 1975 "70s"
3 ef 1985 "80s"
4 gh 1960 "older"
5 ij 1955 "older"
I could use a for-loop checking every single entry in the year-column, but there should be a smarter and faster way using apply or similar functions. Unfortunately, I can't figure that out.
n <- c(1,2,3,4,5)
a <- c("ab", "cd", "ef", "gh", "ij")
y <- c(1974, 1975, 1985, 1960, 1955)
df <- data.frame(numbers = n, any_data = a, year = y, period = NA)
df$period <- factor(df$period, levels =c("70s", "80s", "older"))
for (i in 1:length(df$year)){
if((df$year[i] 1969) && (df$year[i] < 1980)){
df$period[i] <- "70s"
}
# and so on
}
df
That's slow and ugly. Any better ideas?
r apply categories
edited Nov 20 '18 at 12:11
kath
4,250924
asked Nov 20 '18 at 11:59
user9204439user9204439
111
add a comment |
0
That's what I have:
numbers any_data year period
1 ab 1974 <NA
2 cd 1975 <NA
3 ef 1985 <NA
4 gh 1960 <NA
5 ij 1955 <NA
...and that is what I want:
numbers any_data year period
1 ab 1974 "70s"
2 cd 1975 "70s"
3 ef 1985 "80s"
4 gh 1960 "older"
5 ij 1955 "older"
I could use a for-loop checking every single entry in the year-column, but there should be a smarter and faster way using apply or similar functions. Unfortunately, I can't figure that out.
n <- c(1,2,3,4,5)
a <- c("ab", "cd", "ef", "gh", "ij")
y <- c(1974, 1975, 1985, 1960, 1955)
df <- data.frame(numbers = n, any_data = a, year = y, period = NA)
df$period <- factor(df$period, levels =c("70s", "80s", "older"))
for (i in 1:length(df$year)){
if((df$year[i] 1969) && (df$year[i] < 1980)){
df$period[i] <- "70s"
}
# and so on
}
df
That's slow and ugly. Any better ideas?
r apply categories
edited Nov 20 '18 at 12:11
kath
4,250924
asked Nov 20 '18 at 11:59
user9204439user9204439
111
6
Readhelp("cut").
– Roland
Nov 20 '18 at 12:02
If your problem got solved please choose an answer.
– Andre Elrico
Nov 22 '18 at 11:11
add a comment |
0
0
0
That's what I have:
numbers any_data year period
1 ab 1974 <NA
2 cd 1975 <NA
3 ef 1985 <NA
4 gh 1960 <NA
5 ij 1955 <NA
...and that is what I want:
numbers any_data year period
1 ab 1974 "70s"
2 cd 1975 "70s"
3 ef 1985 "80s"
4 gh 1960 "older"
5 ij 1955 "older"
I could use a for-loop checking every single entry in the year-column, but there should be a smarter and faster way using apply or similar functions. Unfortunately, I can't figure that out.
n <- c(1,2,3,4,5)
a <- c("ab", "cd", "ef", "gh", "ij")
y <- c(1974, 1975, 1985, 1960, 1955)
df <- data.frame(numbers = n, any_data = a, year = y, period = NA)
df$period <- factor(df$period, levels =c("70s", "80s", "older"))
for (i in 1:length(df$year)){
if((df$year[i] 1969) && (df$year[i] < 1980)){
df$period[i] <- "70s"
}
# and so on
}
df
That's slow and ugly. Any better ideas?
r apply categories
edited Nov 20 '18 at 12:11
kath
4,250924
asked Nov 20 '18 at 11:59
user9204439user9204439
111
That's what I have:
numbers any_data year period
1 ab 1974 <NA
2 cd 1975 <NA
3 ef 1985 <NA
4 gh 1960 <NA
5 ij 1955 <NA
...and that is what I want:
numbers any_data year period
1 ab 1974 "70s"
2 cd 1975 "70s"
3 ef 1985 "80s"
4 gh 1960 "older"
5 ij 1955 "older"
I could use a for-loop checking every single entry in the year-column, but there should be a smarter and faster way using apply or similar functions. Unfortunately, I can't figure that out.
n <- c(1,2,3,4,5)
a <- c("ab", "cd", "ef", "gh", "ij")
y <- c(1974, 1975, 1985, 1960, 1955)
df <- data.frame(numbers = n, any_data = a, year = y, period = NA)
df$period <- factor(df$period, levels =c("70s", "80s", "older"))
for (i in 1:length(df$year)){
if((df$year[i] 1969) && (df$year[i] < 1980)){
df$period[i] <- "70s"
}
# and so on
}
df
That's slow and ugly. Any better ideas?
r apply categories
r apply categories
edited Nov 20 '18 at 12:11
kath
4,250924
asked Nov 20 '18 at 11:59
user9204439user9204439
111
edited Nov 20 '18 at 12:11
kath
4,250924
asked Nov 20 '18 at 11:59
user9204439user9204439
111
edited Nov 20 '18 at 12:11
kath
4,250924
edited Nov 20 '18 at 12:11
kath
4,250924
edited Nov 20 '18 at 12:11
kath
4,250924
4,250924
asked Nov 20 '18 at 11:59
user9204439user9204439
111
asked Nov 20 '18 at 11:59
user9204439user9204439
111
asked Nov 20 '18 at 11:59
user9204439user9204439
111
111
6
Readhelp("cut").
– Roland
Nov 20 '18 at 12:02
If your problem got solved please choose an answer.
– Andre Elrico
Nov 22 '18 at 11:11
add a comment |
6
Readhelp("cut").
– Roland
Nov 20 '18 at 12:02
If your problem got solved please choose an answer.
– Andre Elrico
Nov 22 '18 at 11:11
6
6
Read
help("cut").– Roland
Nov 20 '18 at 12:02
Read
help("cut").– Roland
Nov 20 '18 at 12:02
If your problem got solved please choose an answer.
– Andre Elrico
Nov 22 '18 at 11:11
If your problem got solved please choose an answer.
– Andre Elrico
Nov 22 '18 at 11:11
add a comment |
1 Answer
1
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oldest
votes
2
a good, readable and general way would be to use dplyr's ?case_when.
df$period <-
(df$y %% 100) %>% {dplyr::case_when( . >= 80 ~ "80s",
. >= 70 ~ "70s",
TRUE ~ "older")}
# [1] "70s" "70s" "80s" "older" "older"
another way using ?ifelse:
dec <- as.integer(substr(y, 3 ,3 ))
df$period <-
ifelse(dec > 6, paste0(dec, "0s"), "older")
# [1] "70s" "70s" "80s" "older" "older"
or ?cut as Roland suggests:
df$period <-
cut((df$y %% 100), breaks=c(-Inf, 70, 80, Inf), labels = c("older", "70s", "80s"), right = FALSE)
#[1] 70s 70s 80s older older
#Levels: older 70s 80s
answered Nov 20 '18 at 12:08
Andre ElricoAndre Elrico
5,70011129
Thanks, Andre! For my problem cut works best.
– user9204439
Nov 21 '18 at 17:16
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
a good, readable and general way would be to use dplyr's ?case_when.
df$period <-
(df$y %% 100) %>% {dplyr::case_when( . >= 80 ~ "80s",
. >= 70 ~ "70s",
TRUE ~ "older")}
# [1] "70s" "70s" "80s" "older" "older"
another way using ?ifelse:
dec <- as.integer(substr(y, 3 ,3 ))
df$period <-
ifelse(dec > 6, paste0(dec, "0s"), "older")
# [1] "70s" "70s" "80s" "older" "older"
or ?cut as Roland suggests:
df$period <-
cut((df$y %% 100), breaks=c(-Inf, 70, 80, Inf), labels = c("older", "70s", "80s"), right = FALSE)
#[1] 70s 70s 80s older older
#Levels: older 70s 80s
answered Nov 20 '18 at 12:08
Andre ElricoAndre Elrico
5,70011129
Thanks, Andre! For my problem cut works best.
– user9204439
Nov 21 '18 at 17:16
add a comment |
2
a good, readable and general way would be to use dplyr's ?case_when.
df$period <-
(df$y %% 100) %>% {dplyr::case_when( . >= 80 ~ "80s",
. >= 70 ~ "70s",
TRUE ~ "older")}
# [1] "70s" "70s" "80s" "older" "older"
another way using ?ifelse:
dec <- as.integer(substr(y, 3 ,3 ))
df$period <-
ifelse(dec > 6, paste0(dec, "0s"), "older")
# [1] "70s" "70s" "80s" "older" "older"
or ?cut as Roland suggests:
df$period <-
cut((df$y %% 100), breaks=c(-Inf, 70, 80, Inf), labels = c("older", "70s", "80s"), right = FALSE)
#[1] 70s 70s 80s older older
#Levels: older 70s 80s
answered Nov 20 '18 at 12:08
Andre ElricoAndre Elrico
5,70011129
Thanks, Andre! For my problem cut works best.
– user9204439
Nov 21 '18 at 17:16
add a comment |
2
2
2
a good, readable and general way would be to use dplyr's ?case_when.
df$period <-
(df$y %% 100) %>% {dplyr::case_when( . >= 80 ~ "80s",
. >= 70 ~ "70s",
TRUE ~ "older")}
# [1] "70s" "70s" "80s" "older" "older"
another way using ?ifelse:
dec <- as.integer(substr(y, 3 ,3 ))
df$period <-
ifelse(dec > 6, paste0(dec, "0s"), "older")
# [1] "70s" "70s" "80s" "older" "older"
or ?cut as Roland suggests:
df$period <-
cut((df$y %% 100), breaks=c(-Inf, 70, 80, Inf), labels = c("older", "70s", "80s"), right = FALSE)
#[1] 70s 70s 80s older older
#Levels: older 70s 80s
answered Nov 20 '18 at 12:08
Andre ElricoAndre Elrico
5,70011129
a good, readable and general way would be to use dplyr's ?case_when.
df$period <-
(df$y %% 100) %>% {dplyr::case_when( . >= 80 ~ "80s",
. >= 70 ~ "70s",
TRUE ~ "older")}
# [1] "70s" "70s" "80s" "older" "older"
another way using ?ifelse:
dec <- as.integer(substr(y, 3 ,3 ))
df$period <-
ifelse(dec > 6, paste0(dec, "0s"), "older")
# [1] "70s" "70s" "80s" "older" "older"
or ?cut as Roland suggests:
df$period <-
cut((df$y %% 100), breaks=c(-Inf, 70, 80, Inf), labels = c("older", "70s", "80s"), right = FALSE)
#[1] 70s 70s 80s older older
#Levels: older 70s 80s
answered Nov 20 '18 at 12:08
Andre ElricoAndre Elrico
5,70011129
edited Nov 20 '18 at 12:52
answered Nov 20 '18 at 12:08
Andre ElricoAndre Elrico
5,70011129
answered Nov 20 '18 at 12:08
Andre ElricoAndre Elrico
5,70011129
answered Nov 20 '18 at 12:08
Andre ElricoAndre Elrico
5,70011129
5,70011129
Thanks, Andre! For my problem cut works best.
– user9204439
Nov 21 '18 at 17:16
add a comment |
Thanks, Andre! For my problem cut works best.
– user9204439
Nov 21 '18 at 17:16
Thanks, Andre! For my problem cut works best.
– user9204439
Nov 21 '18 at 17:16
Thanks, Andre! For my problem cut works best.
– user9204439
Nov 21 '18 at 17:16
add a comment |
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6
Read
help("cut").– Roland
Nov 20 '18 at 12:02
If your problem got solved please choose an answer.
– Andre Elrico
Nov 22 '18 at 11:11