Question

我有一个月份（数字）列表，我将其转换为名称：

fd <- df %>%
  select(product, sales, month) %>%
  mutate(month = month.name[month]) %>%
  filter(!is.na(product), sales!=0) %>%
  group_by(month) %>%
  summarise(sales = sum(sales)) %>%
  collect()

我想对表格进行排序，以便按时间顺序列出月份。如果可能的话，我正在寻找arrange()来自dplyr的解决方案。

以下是fd的结果：

       month   sales
1      April 1306629
2     August 1317986
3   December 1263070
4   February 1493914
5    January 1316889
6       July 1323161
7       June 1331614
8      March 1439019
9        May 1369881
10  November 1256950
11   October 1317647
12 September 1229632

Answer 1

这种方法怎么样：

set.seed(12)
df <- data.frame(month = sample(12), x = LETTERS[1:12])
df
#   month x
#1      1 A
#2      9 B
#3     10 C
#4      3 D
#5      2 E
#6     12 F
#7      8 G
#8      4 H
#9      7 I
#10     5 J
#11    11 K
#12     6 L

library(dplyr)
df %>% 
   mutate(month = factor(month.name[month], levels = month.name)) %>% 
   arrange(month)

#       month x
#1    January A
#2   February E
#3      March D
#4      April H
#5        May J
#6       June L
#7       July I
#8     August G
#9  September B
#10   October C
#11  November K
#12  December F

按照dplyr中的arrange（）按时间顺序排序

1 个答案: