R：是否可以将 2 个变量合并为一个变量？

问题描述 投票：0回答：4

是否可以将 2 个变量合并为一个唯一的变量？

这是我的数据集

`City Name`  `Average amount` `Nominal Difference`   `%`   `Real Wage`    `Real Difference`

Barletta           2457            1007            41.0        27.5              11.3
Caserta            2445            910             37.2        27.4              10.2
Avellino           2363            1016            43.0        26.5              11.4
Lecce              2342            981             41.9        26.2              11.0
Benevento          2335            1157            49.6        26.1              13.0
Isernia            2334            1078            46.2        26.1              12.1
L'Aquila           2324            1010            43.5        26.0              11.3
Catanzaro          2310            1533            66.3        25.9              17.2
Campobasso         2259            1106            49.0        25.3              12.4
Enna               2242            922             41.1        25.1              10.3

我想在名义差异栏中有百分比（带加号和括号内）。

为了更好地解释

我要找的是

`City Name`  `Average amount` `Nominal Difference`   `Real Wage`    `Real Difference`

Barletta           2457            1007 (+41.0%)        27.5              11.3
Caserta            2445            910  (+37.2%)        27.4              10.2
Avellino           2363            1016 (+43.0%)        26.5              11.4
Lecce              2342            981  (+41.9%)        26.2              11.0
Benevento          2335            1157 (+49.6%)        26.1              13.0
Isernia            2334            1078 (+46.2%)        26.1              12.1
L'Aquila           2324            1010 (+43.5%)        26.0              11.3
Catanzaro          2310            1533 (+66.3%)        25.9              17.2
Campobasso         2259            1106 (+49.0%)        25.3              12.4
Enna               2242            922  (+41.1%)        25.1              10.3

我该怎么做？

更新

> dput(f)
structure(list(`City Name` = c("Barletta -Andria-Trani", "Caserta", 
"Avellino", "Lecce", "Benevento", "Isernia", "L'Aquila", "Catanzaro", 
"Campobasso", "Enna"), `Average amount` = c(2456.92, 2444.58, 
2363.48, 2341.57, 2334.63, 2334.01, 2323.97, 2310.46, 2259.03, 
2242.38), `Nominal Difference` = c(1006.8, 909.62, 1016.28, 980.7, 
1157.25, 1077.51, 1010.32, 1532.79, 1106.31, 922.35), `%` = c(40.97, 
37.2, 42.99, 41.88, 49.56, 46.16, 43.47, 66.34, 48.97, 41.13), 
    `Real Wage` = c(27.51, 27.37, 26.46, 26.22, 26.14, 26.13, 
    26.02, 25.87, 25.29, 25.11), `Real Difference` = c(11.27, 
    10.18, 11.38, 10.98, 12.95, 12.06, 11.31, 17.16, 12.38, 10.32
    )), row.names = c(NA, -10L), class = c("tbl_df", "tbl", "data.frame"
))
r dataframe variables dplyr merge
4个回答
1
投票

来自我的评论：

with(dataset, sprintf("%.0f (%+0.1f%%)", `Nominal Difference`, `Real Wage`)) 
#  [1] "1007 (+27.5%)" "910 (+27.4%)"  "1016 (+26.5%)" "981 (+26.2%)"  "1157 (+26.1%)" "1078 (+26.1%)" "1010 (+26.0%)"
#  [8] "1533 (+25.9%)" "1106 (+25.3%)" "922 (+25.1%)"

数据

dataset <- structure(list("City Name" = c("Barletta", "Caserta", "Avellino", "Lecce", "Benevento", "Isernia", "L'Aquila", "Catanzaro", "Campobasso", "Enna"), "Average amount" = c(2457L, 2445L, 2363L, 2342L, 2335L, 2334L, 2324L, 2310L, 2259L, 2242L), "Nominal Difference" = c(1007L, 910L, 1016L, 981L, 1157L, 1078L, 1010L, 1533L, 1106L, 922L), "%" = c(41, 37.2, 43, 41.9, 49.6, 46.2, 43.5, 66.3, 49, 41.1), "Real Wage" = c(27.5, 27.4, 26.5, 26.2, 26.1, 26.1, 26, 25.9, 25.3, 25.1), "Real Difference" = c(11.3, 10.2,  11.4, 11, 13, 12.1, 11.3, 17.2, 12.4, 10.3)), class = "data.frame", row.names = c(NA, -10L))
0
投票

我们可以在

paste0()
中使用
mutate()
并用
sprintf()
调整小数点：

library(dplyr)

mydat <- tribble(~ `City Name`, ~ `Nominal Difference`,   ~`%`,
"Barletta", 1007, 41.0,
"Caserta",  910, 37.2
)

mydat %>% 
  mutate(`Nominal Difference` = paste0(`Nominal Difference`, sprintf(" (+%.1f)",`%`)))
#> # A tibble: 2 x 3
#>   `City Name` `Nominal Difference`   `%`
#>   <chr>       <chr>                <dbl>
#> 1 Barletta    1007 (+41.0)          41  
#> 2 Caserta     910 (+37.2)           37.2

创建于 2023-02-27 由 reprex 包 (v2.0.1)

0
投票

我想你可以使用

unite
，例如

df %>%
  mutate(perc = sprintf("(+%.1f%%)", perc)) %>%
  unite(val, val:perc, remove = TRUE, sep = " ")

输出像

   id         val
1   1 68 (+68.7%)
2   2 39 (+38.4%)
3   3  1 (+77.0%)
4   4 34 (+49.8%)
5   5 87 (+71.8%)
6   6 43 (+99.2%)
7   7 14 (+38.0%)
8   8 82 (+77.7%)
9   9 59 (+93.5%)
10 10 51 (+21.2%)

虚拟数据

set.seed(1)
df <- data.frame(
  id = 1:10,
  val = sample(100, 10),
  perc = 100 * runif(10)
)

# > df
#    id val     perc
# 1   1  68 68.70228
# 2   2  39 38.41037
# 3   3   1 76.98414
# 4   4  34 49.76992
# 5   5  87 71.76185
# 6   6  43 99.19061
# 7   7  14 38.00352
# 8   8  82 77.74452
# 9   9  59 93.47052
# 10 10  51 21.21425
0
投票
df %>%
  mutate(across(percent, ~ str_c("(", .x, "%)"))) %>% 
  unite("nominal_difference", c(nominal_difference, percent), sep = " ") 

# A tibble: 10 × 5
   city_name  average_amount nominal_difference real_wage real_difference
   <chr>               <dbl> <chr>                  <dbl>           <dbl>
 1 Barletta             2457 1007 (41%)              27.5            11.3
 2 Caserta              2445 910 (37.2%)             27.4            10.2
 3 Avellino             2363 1016 (43%)              26.5            11.4
 4 Lecce                2342 981 (41.9%)             26.2            11  
 5 Benevento            2335 1157 (49.6%)            26.1            13  
 6 Isernia              2334 1078 (46.2%)            26.1            12.1
 7 L'Aquila             2324 1010 (43.5%)            26              11.3
 8 Catanzaro            2310 1533 (66.3%)            25.9            17.2
 9 Campobasso           2259 1106 (49%)              25.3            12.4
10 Enna                 2242 922 (41.1%)             25.1            10.3
最新问题
© www.soinside.com 2019 - 2024. All rights reserved.