在指定的列上操作numcolwise

Question

我正在使用大型越野面板数据。这是我的数据示例：

df <- structure(list(country = c("Argentina", "Argentina", "Argentina", 
"Argentina", "Argentina", "Argentina", "Argentina", "Argentina", 
"Argentina", "Argentina", "Argentina", "Argentina", "Argentina", 
"Argentina", "Argentina", "Brazil", "Brazil", "Brazil", "Brazil", 
"Brazil", "Brazil", "Brazil", "Brazil", "Brazil", "Brazil", "Brazil", 
"Brazil", "Brazil", "Brazil", "Brazil"), year = c(1991, 1992, 
1993, 1994, 1995, 1996, 1997, 1998, 1999, 2000, 2001, 2002, 2003, 
2004, 2005, 1991, 1992, 1993, 1994, 1995, 1996, 1997, 1998, 1999, 
2000, 2001, 2002, 2003, 2004, 2005), lnunderval = c(-0.942018220566855, 
-0.885848248127534, -0.766349222095516, -0.690487190951407, -0.521023028925771, 
-0.288557433912095, -0.351488637772915, -0.393048184068511, -0.444123691025518, 
-0.512425182981147, -0.541182815398097, 0.379018666505875, 0.291852440172936, 
0.291407056285245, 0.221426753100227, -0.120418577004832, 0.00467960055625634, 
-0.0190735963658737, -0.239570582118898, -0.316748349307701, 
-0.205418347557874, -0.301707274202926, -0.346946676711871, -0.0528811487098006, 
-0.178001370772517, -0.0404491572081528, 0.0898307782259906, 
0.0835291098039626, 0.0349739055576117, -0.187321483795299), 
    manu_GDP = c(24.3864490932335, 21.8591315586603, 18.2399115325496, 
    17.8190917106899, 17.2467521148076, 17.5357232920479, 18.227905749866, 
    17.8379584760908, 16.9615250614589, 16.4942719439838, 16.0932258763829, 
    20.347773913878, 22.4867505875749, 18.9370136214371, 18.340415936715, 
    21.8391379495813, 23.3085986320751, 26.0497364463813, 23.7212337008806, 
    14.5422791544751, 13.0671912367218, 13.0186253732125, 12.1551371940101, 
    12.3085333305115, 13.134659593552, 13.0895379354001, 12.3569626673735, 
    14.4507645630532, 15.0995301563871, 14.7382811342998), income = c("Upper middle income", 
    "Upper middle income", "Upper middle income", "Upper middle income", 
    "Upper middle income", "Upper middle income", "Upper middle income", 
    "Upper middle income", "Upper middle income", "Upper middle income", 
    "Upper middle income", "Upper middle income", "Upper middle income", 
    "Upper middle income", "Upper middle income", "Upper middle income", 
    "Upper middle income", "Upper middle income", "Upper middle income", 
    "Upper middle income", "Upper middle income", "Upper middle income", 
    "Upper middle income", "Upper middle income", "Upper middle income", 
    "Upper middle income", "Upper middle income", "Upper middle income", 
    "Upper middle income", "Upper middle income"), period = structure(c(1L, 
    1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 1L, 
    1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L), .Label = c("(1990,1995]", 
    "(1995,2000]", "(2000,2005]"), class = "factor")), row.names = c(NA, 
-30L), class = c("tbl_df", "tbl", "data.frame"))

我使用剪切和ddply函数如下创建了我的变量的五年非重叠平均值。

df$period <- cut(df$year, seq(1990, 2005, 5)) #this periodizes data
df <- ddply(df, .(country, period), numcolwise(mean)) 

该代码的问题在于，命名收入的非数字列丢失。我已经尝试了以下操作，但它不起作用。

df <- ddply(df, .(country, period), numcolwise(mean,.(lnunderval, manu_GDP))) 
Error in mean.default(X[[i]], ...) : 'trim' must be numeric of length one

我想最终数据集包含未平均的非数字列。是否有一种方法可以在指定的一组列上应用NumColwise函数？

我希望最终的输出看起来像这样：

structure(list(country = c("Argentina", "Argentina", "Argentina", 
"Brazil", "Brazil", "Brazil"), period = structure(c(1L, 2L, 3L, 
1L, 2L, 3L), .Label = c("(1990,1995]", "(1995,2000]", "(2000,2005]"
), class = "factor"), year = c(1993, 1998, 2003, 1993, 1998, 
2003), lnunderval = c(-0.761145182133417, -0.397928625952037, 
0.128504420133237, -0.13822630084821, -0.216990963590998, -0.00388736948317731
), manu_GDP = c(19.9102672019882, 17.4114769046895, 19.2410359871976, 
21.8921971766787, 12.7368293456016, 13.9470152913027), income = c("Upper middle income", 
"Upper middle income", "Upper middle income", "Upper middle income", 
"Upper middle income", "Upper middle income")), class = "data.frame", row.names = c(NA, 
-6L))

Answer 1

我们可以使用dplyr，它在 to 的 多个具有不同功能的列块中更灵活

library(dplyr)
df %>%
  group_by(country, period) %>%
  summarise(year = last(year), income = list(unique(income[!is.na(income)])), 
    across(c(lnunderval, manu_GDP), mean), .groups = 'drop')