dplyr：根据聚合函数结果过滤行 [英] dplyr: filter rows according to aggregated function result

问题描述

我有一个表格列表（金额，年份和月份），我想过滤与完整年份相对应的行。即我想忽略下面给出的示例数据框的最后4行，它指的是2015，并得到其余的60.可以用一个dplyr命令来做到这一点吗？

I have a table listing (amount, year and month) and I want to filter the rows corresponding to complete years. I.e. I want to ommit the last 4 rows of the sample dataframe I give below, that refer to 2015, and get the rest 60. Is it possible to do that with a single dplyr command?

我尝试过：

df %>%
    group_by(year) %>%
    tally() %>%
    filter (n==12) %>%
    ungroup() 

但我想ungroup做的不同于我想要的东西。可以通过一个dplyr命令来做到这一点吗？

but I guess ungroup does something different than what I want. Is it possible to do that with a single dplyr command?

df <- structure(list(amount = c(16365, 31850, 32230, 34177.75, 27900, 
29650, 28846, 27300, 37115.31, 34130.38, 39676.1, 47244.44, 3500, 
25425.48, 22628.43, 30822.86, 30100, 41567.13, 25400, 23125, 
40073.75, 16505.82, 17770, 38406.03, 1528.25, 23475.77, 29869.69, 
17020, 19270, 13085.47, 10607.48, 7800, 15220, 15260, 17580, 
25094.66, 3908.74, 8150, 25055.89, 19690.65, 12445.4, 10347.39, 
7645.39, 49300, 8690, 13660, 16510, 34457.08, 522.68, 10202, 
18900, 25027.1, 24956.42, 23259, 32743, 37226, 32697, 32258, 
31336.67, 36135.81, 4389.26, 12450, 46220.43, 36770.7), year = c("2010", 
"2010", "2010", "2010", "2010", "2010", "2010", "2010", "2010", 
"2010", "2010", "2010", "2011", "2011", "2011", "2011", "2011", 
"2011", "2011", "2011", "2011", "2011", "2011", "2011", "2012", 
"2012", "2012", "2012", "2012", "2012", "2012", "2012", "2012", 
"2012", "2012", "2012", "2013", "2013", "2013", "2013", "2013", 
"2013", "2013", "2013", "2013", "2013", "2013", "2013", "2014", 
"2014", "2014", "2014", "2014", "2014", "2014", "2014", "2014", 
"2014", "2014", "2014", "2015", "2015", "2015", "2015"), month = c("01", 
"02", "03", "04", "05", "06", "07", "08", "09", "10", "11", "12", 
"01", "02", "03", "04", "05", "06", "07", "08", "09", "10", "11", 
"12", "01", "02", "03", "04", "05", "06", "07", "08", "09", "10", 
"11", "12", "01", "02", "03", "04", "05", "06", "07", "08", "09", 
"10", "11", "12", "01", "02", "03", "04", "05", "06", "07", "08", 
"09", "10", "11", "12", "01", "02", "03", "04")), .Names = c("amount", 
"year", "month"), class = c("tbl_df", "data.frame"), row.names = c(NA, 
-64L))

推荐答案

tally（）相当于 summarize（n = n（）） 。但是，在这种情况下，您希望保留数据帧的原始行，但要进行过滤，以便删除不完整年份的行。 @ AndresT的答案可以正常工作，但是您也可以更简洁地进行操作，而无需创建列来计算每个组的行数的中间步骤：

tally() is the equivalent of summarise(n=n()). However, in this case you want to keep the original rows of the data frame, but filtered so that rows that are part of incomplete years are removed. @AndresT's answer will work fine, but you can also do it more concisely without an intermediate step of creating a column to count the number of rows for each group:

df %>% group_by(year) %>% filter(n()==12)

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