x = list(data.frame(age = c(1:4),period = c(2000:2003)),
data.frame(age = c(5:8),period = c(1998:2001)),
data.frame(age = c(11:19),period = c(1990:1998)))
map2(x, x$period, ~cbind(.x, difference = .y-.x$age))
result:
> map2(x, x$period, ~cbind(.x, difference = .y-.x$age))
list()
Is it possible to map the function by using the elements within the same dataframe?
In your context x$period is NULL since x is the list of dataframes and it has no attribute "period". I think you want to access the period column within each unnammed dataframe in the list. I would just use map which will pass along each dataframe in the list, which you can then manipulate in the function to access each column without having to explicitly pass it.
library(purrr)
library(dplyr)
x = list(data.frame(age = c(1:4),period = c(2000:2003)),
data.frame(age = c(5:8),period = c(1998:2001)),
data.frame(age = c(11:19),period = c(1990:1998)))
#Original attempt
result <- map2(x, x$period, ~cbind(.x, difference = .y-.x$age))
result
#> list()
#My solution
result2 <- map(x, function(df) cbind(df, difference = df$period - df$age))
result2
#> [[1]]
#> age period difference
#> 1 1 2000 1999
#> 2 2 2001 1999
#> 3 3 2002 1999
#> 4 4 2003 1999
#>
#> [[2]]
#> age period difference
#> 1 5 1998 1993
#> 2 6 1999 1993
#> 3 7 2000 1993
#> 4 8 2001 1993
#>
#> [[3]]
#> age period difference
#> 1 11 1990 1979
#> 2 12 1991 1979
#> 3 13 1992 1979
#> 4 14 1993 1979
#> 5 15 1994 1979
#> 6 16 1995 1979
#> 7 17 1996 1979
#> 8 18 1997 1979
#> 9 19 1998 1979
#A more readable solution using dplyr
result3 <- map(x, function(df) df %>% mutate(difference = period - age))
result3
#> [[1]]
#> age period difference
#> 1 1 2000 1999
#> 2 2 2001 1999
#> 3 3 2002 1999
#> 4 4 2003 1999
#>
#> [[2]]
#> age period difference
#> 1 5 1998 1993
#> 2 6 1999 1993
#> 3 7 2000 1993
#> 4 8 2001 1993
#>
#> [[3]]
#> age period difference
#> 1 11 1990 1979
#> 2 12 1991 1979
#> 3 13 1992 1979
#> 4 14 1993 1979
#> 5 15 1994 1979
#> 6 16 1995 1979
#> 7 17 1996 1979
#> 8 18 1997 1979
#> 9 19 1998 1979
Created on 2023-02-02 with reprex v2.0.2
Related
So I have a data table of 5000 firms, each firm is assigned a numerical value ("id") which is 1 for the first firm, 2 for the second ...
Here is my table with only the profit variable :
|id | year | profit
|:----| :----| :----|
|1 |2001 |-0.4
|1 |2002 |-0.89
|2 |2001 |1.89
|2 |2002 |2.79
Each firm is expressed twice, one line specifies the data in 2001 and the second in 2002 (the "id" value being the same on both lines because it is the same firm one year apart).
How to calculate the annual rate of change of each firm ("id") between 2001 and 2002 ?
I'm really new to R and I don't see where to start? Separate the 2001 and 2002 data?
I did this :
years <- sort(unique(group$year))years
And I also found this on the internet but with no success :
library(dplyr)
res <-
group %>%
arrange(id,year) %>%
group_by(id) %>%
mutate(evol_rate = ("group$year$2002" / lag("group$year$2001") - 1) * 100) %>%
ungroup()
Thank you very much
From what you've written, I take it that you want to calculate the formula for ROC for the profit values of 2001 and 2002:
ROC=(current_value/previous_value − 1) ∗ 100
To accomplish this, I suggest tidyr::pivot_wider() which reshapes your dataframe from long to wide format (see: https://r4ds.had.co.nz/tidy-data.html#pivoting).
Code:
require(tidyr)
require(dplyr)
id <- sort(rep(seq(1,250, 1), 2))
year <- rep(seq(2001, 2002, 1), 500)
value <- sample(500:2000, 500)
df <- data.frame(id, year, value)
head(df, 10)
#> id year value
#> 1 1 2001 856
#> 2 1 2002 1850
#> 3 2 2001 1687
#> 4 2 2002 1902
#> 5 3 2001 1728
#> 6 3 2002 1773
#> 7 4 2001 691
#> 8 4 2002 1691
#> 9 5 2001 1368
#> 10 5 2002 893
df_wide <- df %>%
pivot_wider(names_from = year,
names_prefix = "profit_",
values_from = value,
values_fn = mean)
res <- df_wide %>%
mutate(evol_rate = (profit_2002/profit_2001-1)*100) %>%
round(2)
head(res, 10)
#> # A tibble: 10 x 4
#> id profit_2001 profit_2002 evol_rate
#> <dbl> <dbl> <dbl> <dbl>
#> 1 1 856 1850 116.
#> 2 2 1687 1902 12.7
#> 3 3 1728 1773 2.6
#> 4 4 691 1691 145.
#> 5 5 1368 893 -34.7
#> 6 6 883 516 -41.6
#> 7 7 1280 1649 28.8
#> 8 8 1579 1383 -12.4
#> 9 9 1907 1626 -14.7
#> 10 10 1227 1134 -7.58
If you want to do it without reshaping your data into a wide format you can use
library(tidyverse)
id <- sort(rep(seq(1,250, 1), 2))
year <- rep(seq(2001, 2002, 1), 500)
value <- sample(500:2000, 500)
df <- data.frame(id, year, value)
df %>% head(n = 10)
#> id year value
#> 1 1 2001 1173
#> 2 1 2002 1648
#> 3 2 2001 1560
#> 4 2 2002 1091
#> 5 3 2001 1736
#> 6 3 2002 667
#> 7 4 2001 1840
#> 8 4 2002 1202
#> 9 5 2001 1597
#> 10 5 2002 1797
new_df <- df %>%
group_by(id) %>%
mutate(ROC = ((value / lag(value) - 1) * 100))
new_df %>% head(n = 10)
#> # A tibble: 10 × 4
#> # Groups: id [5]
#> id year value ROC
#> <dbl> <dbl> <int> <dbl>
#> 1 1 2001 1173 NA
#> 2 1 2002 1648 40.5
#> 3 2 2001 1560 NA
#> 4 2 2002 1091 -30.1
#> 5 3 2001 1736 NA
#> 6 3 2002 667 -61.6
#> 7 4 2001 1840 NA
#> 8 4 2002 1202 -34.7
#> 9 5 2001 1597 NA
#> 10 5 2002 1797 12.5
This groups the data by id and then uses lag to compare the current year to the year prior
I have this data:
data <- data.frame(id_pers=c(4102,13102,27101,27102,28101,28102, 42101,42102,56102,73102,74103,103104,117103,117104,117105),
birthyear=c(1992,1994,1993,1992,1995,1999,2000,2001,2000, 1994, 1999, 1978, 1986, 1998, 1999))
I want to group the different persons by familys in a new column, so that persons 27101,27102 (siblings) are group/family 1 and 42101,42102 are in group 2, 117103,117104,117105 are in group 3 so on.
Person "4102" has no siblings and should be a NA in the new column.
It is always the case that 2 or more persons are siblings if the ID's are not further apart than a maximum of 6 numbers.
I have a far larger dataset with over 3000 rows. How could I do it the most efficient way?
You can use round with digits = -1 (or -2) if you have id_pers that goes above 10 observations per family. If you want the id to be integers from 1; you can use cur_group_id:
library(dplyr)
data %>%
group_by(fam_id = round(id_pers - 5, digits = -1)) %>%
mutate(fam_gp = cur_group_id())
output
# A tibble: 15 × 3
# Groups: fam_id [10]
id_pers birthyear fam_id fam_gp
<dbl> <dbl> <dbl> <int>
1 4102 1992 4100 1
2 13102 1994 13100 2
3 27101 1993 27100 3
4 27102 1992 27100 3
5 28101 1995 28100 4
6 28106 1999 28100 4
7 42101 2000 42100 5
8 42102 2001 42100 5
9 56102 2000 56100 6
10 73102 1994 73100 7
11 74103 1999 74100 8
12 103104 1978 103100 9
13 117103 1986 117100 10
14 117104 1998 117100 10
15 117105 1999 117100 10
It looks like we can the 1000s digit (and above) to delineate groups.
library(dplyr)
data %>%
mutate(
famgroup = trunc(id_pers/1000),
famgroup = match(famgroup, unique(famgroup))
)
# id_pers birthyear famgroup
# 1 4102 1992 1
# 2 13102 1994 2
# 3 27101 1993 3
# 4 27102 1992 3
# 5 28101 1995 4
# 6 28102 1999 4
# 7 42101 2000 5
# 8 42102 2001 5
# 9 56102 2000 6
# 10 73102 1994 7
# 11 74103 1999 8
# 12 103104 1978 9
# 13 117103 1986 10
# 14 117104 1998 10
# 15 117105 1999 10
I'd like to preserve the proper yearly index names as I recast my data from wide to long.
dt = data.table(country = c(1,2,3,4,5), gdp_1990 = rnorm(5), gdp_1991 = rnorm(5), gdp_1992 = rnorm(5),
unemp_1990 = rnorm(5), unemp_1991 = rnorm(5), unemp_1992 = rnorm(5))
melt(dt, id = 'country', measure = patterns(gdp = '^gdp_', unemp = '^unemp_'), variable.name = 'year')
Desired Output:
country year gdp unemp
1: 1 1990 0.856957066 -1.42947033
2: 2 1990 -1.765995901 1.38170009
3: 3 1990 -0.298302521 -0.54070574
4: 4 1990 -0.919421829 -0.17552704
5: 5 1990 -0.189133135 1.18923546
6: 1 1991 -1.248963381 -0.10467153
7: 2 1991 -0.800931881 0.03589986
Actual Output:
country year gdp unemp
1: 1 1 0.856957066 -1.42947033
2: 2 1 -1.765995901 1.38170009
3: 3 1 -0.298302521 -0.54070574
4: 4 1 -0.919421829 -0.17552704
5: 5 1 -0.189133135 1.18923546
6: 1 2 -1.248963381 -0.10467153
7: 2 2 -0.800931881 0.03589986
With data.table (dev version - 1.14.3) we can use measure with sep as documented in ?measure
measure(..., sep, pattern, cols, multiple.keyword="value.name")
library(data.table)
melt(dt, measure.vars = measure(value.name, year, sep = "_"))
-output
country year gdp unemp
<num> <char> <num> <num>
1: 1 1990 -1.275041172 -0.75524345
2: 2 1990 1.979629503 -1.14636877
3: 3 1990 0.062272176 1.16928396
4: 4 1990 -0.210106506 -0.66517069
5: 5 1990 -1.089511759 -1.79322014
6: 1 1991 0.460566878 0.61720109
7: 2 1991 0.183378182 -0.01628616
8: 3 1991 -0.647174381 1.14346303
9: 4 1991 0.008846161 0.05223651
10: 5 1991 -0.039701540 1.40848433
11: 1 1992 0.328204416 1.44638191
12: 2 1992 -1.359373393 1.33391755
13: 3 1992 -0.538430362 -0.26828537
14: 4 1992 0.424461192 -0.32107074
15: 5 1992 -0.338010393 -0.19920506
Using tidyr::pivot_longer we can use names_sep = "_" to split the names into the variable and year. In names_to, use the special string ".value" to specify that you want multiple columns created from the gdp and unemp columns:
tidyr::pivot_longer(dt, -1, names_sep = "_", names_to = c(".value", "year"))
#> # A tibble: 15 x 4
#> country year gdp unemp
#> <dbl> <chr> <dbl> <dbl>
#> 1 1 1990 -0.324 -1.12
#> 2 1 1991 0.307 -1.64
#> 3 1 1992 -0.0569 -1.49
#> 4 2 1990 0.0602 -0.751
#> 5 2 1991 -1.54 0.450
#> 6 2 1992 -1.91 -1.08
#> 7 3 1990 -0.589 2.09
#> 8 3 1991 -0.301 -0.0186
#> 9 3 1992 1.18 1.00
#> 10 4 1990 0.531 0.0174
#> 11 4 1991 -0.528 -0.318
#> 12 4 1992 -1.66 -0.621
#> 13 5 1990 -1.52 -1.29
#> 14 5 1991 -0.652 -0.929
#> 15 5 1992 -0.464 -1.38
I have a data frame with three columns: birth_year, death_year, gender.
I have to calculate total alive male and female population for every year in a given range (1950:1980).
The data frame looks like this:
birth_year death_year gender
1934 1988 male
1922 1993 female
1890 1966 male
1901 1956 male
1946 2009 female
1909 1976 female
1899 1945 male
1887 1949 male
1902 1984 female
The person is alive in year x if death_year > x & birth year <= x
The output I am looking for is something like this:
year male female
1950 3 4
1951 2 3
1952 4 3
1953 4 5
.
.
1980 6 3
Thanks!
Does this work:
library(tidyr)
library(purrr)
library(dplyr)
df %>% mutate(year = map2(1950,1980, seq)) %>% unnest(year) %>%
mutate(isalive = case_when(year >= birth_year & year < death_year ~ 1, TRUE ~ 0)) %>%
group_by(year, gender) %>% summarise(alive = sum(isalive)) %>%
pivot_wider(names_from = gender, values_from = alive) %>% print( n = 50)
`summarise()` regrouping output by 'year' (override with `.groups` argument)
# A tibble: 31 x 3
# Groups: year [31]
year female male
<int> <dbl> <dbl>
1 1950 4 3
2 1951 4 3
3 1952 4 3
4 1953 4 3
5 1954 4 3
6 1955 4 3
7 1956 4 2
8 1957 4 2
9 1958 4 2
10 1959 4 2
11 1960 4 2
12 1961 4 2
13 1962 4 2
14 1963 4 2
15 1964 4 2
16 1965 4 2
17 1966 4 1
18 1967 4 1
19 1968 4 1
20 1969 4 1
21 1970 4 1
22 1971 4 1
23 1972 4 1
24 1973 4 1
25 1974 4 1
26 1975 4 1
27 1976 3 1
28 1977 3 1
29 1978 3 1
30 1979 3 1
31 1980 3 1
Data used:
df
# A tibble: 9 x 3
birth_year death_year gender
<dbl> <dbl> <chr>
1 1934 1988 male
2 1922 1993 female
3 1890 1966 male
4 1901 1956 male
5 1946 2009 female
6 1909 1976 female
7 1899 1945 male
8 1887 1949 male
9 1902 1984 female
Here's a simple base R solution. Summing a logical vector will get you your count of alive or dead because TRUE is 1 and FALSE is 0.
number_alive <- function(range, df){
sapply(range, function(x) sum((df$death_year > x) & (df$birth_year <= x)))
}
output <- data.frame('year' = 1950:1980,
'female' = number_alive(1950:1980, df[df$gender == 'female']),
'male' = number_alive(1950:1980, df[df$gender == 'male']))
# year female male
# 1 1950 4 3
# 2 1951 4 3
# 3 1952 4 3
# 4 1953 4 3
# 5 1954 4 3
# 6 1955 4 3
# 7 1956 4 2
# 8 1957 4 2
# 9 1958 4 2
# 10 1959 4 2
# 11 1960 4 2
# 12 1961 4 2
# 13 1962 4 2
# 14 1963 4 2
# 15 1964 4 2
# 16 1965 4 2
# 17 1966 4 1
# 18 1967 4 1
# 19 1968 4 1
# 20 1969 4 1
# 21 1970 4 1
# 22 1971 4 1
# 23 1972 4 1
# 24 1973 4 1
# 25 1974 4 1
# 26 1975 4 1
# 27 1976 3 1
# 28 1977 3 1
# 29 1978 3 1
# 30 1979 3 1
# 31 1980 3 1
This approach uses an ifelse to determine if alive (1) or dead (0).
Data:
df <- "birth_year death_year gender
1934 1988 male
1922 1993 female
1890 1966 male
1901 1956 male
1946 2009 female
1909 1976 female
1899 1945 male
1887 1949 male
1902 1984 female"
df <- read.table(text = df, header = TRUE)
Code:
library(dplyr)
library(tidyr)
library(tibble)
library(purrr)
df %>%
mutate(year = map2(1950,1980, seq)) %>%
unnest(year) %>%
select(year, birth_year, death_year, gender) %>%
mutate(
alive = ifelse(year >= birth_year & year <= death_year, 1, 0)
) %>%
group_by(year, gender) %>%
summarise(
is_alive = sum(alive)
) %>%
pivot_wider(
names_from = gender,
values_from = is_alive
) %>%
select(year, male, female)
Output:
#> # A tibble: 31 x 3
#> # Groups: year [31]
#> year male female
#> <int> <dbl> <dbl>
#> 1 1950 3 4
#> 2 1951 3 4
#> 3 1952 3 4
#> 4 1953 3 4
#> 5 1954 3 4
#> 6 1955 3 4
#> 7 1956 3 4
#> 8 1957 2 4
#> 9 1958 2 4
#> 10 1959 2 4
#> # … with 21 more rows
Created on 2020-11-11 by the reprex package (v0.3.0)
I'm struggling to understand exactly how to compute a deflation factor for wages in a panel based on inflation.
I've teh R example below to help me illustrate the issue.
In Wooldridge (2009:452) Introductory Econometrics, 5th ed., he creates a deflation factor by dividing 107.6 by 65.2, i.e. 107.6/65.2 ≈ 1.65, but I can't figure out to to apply this to my own panel data. Wooldridge only mentions the deflation factor in passing.
Say I have a mini panel with two people, Jane and Tom, staring from 2006/2009 and running until 2015 with their yearly wage,
# install.packages(c("dplyr"), dependencies = TRUE)
library(dplyr)
set.seed(2)
tbl <- tibble(id = rep(c('Jane', 'Tom'), c(7, 10)),
yr = c(2009:2015, 2006:2015),
wg = c(rnorm(7, mean=5.1*10^4, sd=9), rnorm(10, 4*10^4, 12))
); tbl
#> A tibble: 17 x 3
#> id yr wg
#> <chr> <int> <dbl>
#> 1 Jane 2009 50991.93
#> 2 Jane 2010 51001.66
#> 3 Jane 2011 51014.29
#> 4 Jane 2012 50989.83
#> 5 Jane 2013 50999.28
#> 6 Jane 2014 51001.19
#> 7 Jane 2015 51006.37
#> 8 Tom 2006 39997.12
#> 9 Tom 2007 40023.81
#> 10 Tom 2008 39998.33
#> 11 Tom 2009 40005.01
#> 12 Tom 2010 40011.78
#> 13 Tom 2011 39995.29
#> 14 Tom 2012 39987.52
#> 15 Tom 2013 40021.39
#> 16 Tom 2014 39972.27
#> 17 Tom 2015 40010.54
I now get the consumer price index (CPI) (using this answer)
# install.packages(c("Quandl"), dependencies = TRUE)
CPI00to16 <- Quandl::Quandl("FRED/CPIAUCSL", collapse="annual",
start_date="2000-01-01", end_date="2016-01-01")
as_tibble(CPI00to16)
#> # A tibble: 17 x 2
#> Date Value
#> <date> <dbl>
#> 1 2016-12-31 238.106
#> 2 2015-12-31 237.846
#> 3 2014-12-31 236.290
#> 4 2013-12-31 234.723
#> 5 2012-12-31 231.221
#> 6 2011-12-31 227.223
#> 7 2010-12-31 220.472
#> 8 2009-12-31 217.347
#> 9 2008-12-31 211.398
#> 10 2007-12-31 211.445
#> 11 2006-12-31 203.100
#> 12 2005-12-31 198.100
#> 13 2004-12-31 191.700
#> 14 2003-12-31 185.500
#> 15 2002-12-31 181.800
#> 16 2001-12-31 177.400
#> 17 2000-12-31 174.600
my question is how do I deflate Jane and Tom's wages cf. Wooldridge 2009 selecting 2015 as the baseline year?
update; following MrSmithGoesToWashington’s comment below.
CPI00to16$yr <- as.numeric(format(CPI00to16$Date,'%Y'))
CPI00to16 <- mutate(CPI00to16, deflation_factor = CPI00to16[2,2]/Value)
df <- tbl %>% inner_join(as_tibble(CPI00to16[,3:4]), by = "yr")
df <- mutate(df, wg_defl = deflation_factor*wg, wg_diff = wg_defl-wg)
df
#> # A tibble: 17 x 6
#> id yr wg deflation_factor wg_defl wg_diff
#> <chr> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 Jane 2009 50991.93 1.094315 55801.21 4809.2844
#> 2 Jane 2010 51001.66 1.078804 55020.78 4019.1176
#> 3 Jane 2011 51014.29 1.046751 53399.28 2384.9910
#> 4 Jane 2012 50989.83 1.028652 52450.80 1460.9728
#> 5 Jane 2013 50999.28 1.013305 51677.83 678.5477
#> 6 Jane 2014 51001.19 1.006585 51337.04 335.8494
#> 7 Jane 2015 51006.37 1.000000 51006.37 0.0000
#> 8 Tom 2006 39997.12 1.171078 46839.76 6842.6394
#> 9 Tom 2007 40023.81 1.124860 45021.18 4997.3691
#> 10 Tom 2008 39998.33 1.125110 45002.53 5004.1909
#> 11 Tom 2009 40005.01 1.094315 43778.07 3773.0575
#> 12 Tom 2010 40011.78 1.078804 43164.86 3153.0747
#> 13 Tom 2011 39995.29 1.046751 41865.12 1869.8369
#> 14 Tom 2012 39987.52 1.028652 41133.26 1145.7322
#> 15 Tom 2013 40021.39 1.013305 40553.87 532.4863
#> 16 Tom 2014 39972.27 1.006585 40235.49 263.2225
#> 17 Tom 2015 40010.54 1.000000 40010.54 0.0000