Question

我有这样的问题。我有一个像：

这样的数据库

Province       cases        year      month 
Newyork         10          2000         1
Newyork         20          2000         2
Newyork         30          2000         3
Newyork         40          2000         4
Los Angeles     30          2000         1
Los Angeles     40          2000         2
Los Angeles     50          2000         3
Los Angeles     60          2000         4

20年来和许多省份的非常大的数据。如何重新组合我的数据以获得这样的一系列时间：

Province      cases.at.1.2000  cases.at.2.2000  cases.at.3.2000  cases.at.4.2000  
Newyork             10               20                30               40
Los Angeles         30               40                50               60

Answer 1

只需使用dcast包中的reshape2：

library(reshape2)

dcast(df, Province~month+year, value.var='cases')
#    Province 1_2000 2_2000 3_2000 4_2000
#1 LosAngeles     30     40     50     60
#2    Newyork     10     20     30     40

数据：

df=structure(list(Province = c("Newyork", "Newyork", "Newyork", "Newyork", "LosAngeles", "LosAngeles", "LosAngeles", "LosAngeles" ), cases = c(10L, 20L, 30L, 40L, 30L, 40L, 50L, 60L), year = c(2000L, 2000L, 2000L, 2000L, 2000L, 2000L, 2000L, 2000L), month = c(1L, 2L, 3L, 4L, 1L, 2L, 3L, 4L)), .Names = c("Province", "cases", "year", "month"), class = "data.frame", row.names = c(NA, -8L ))

编辑：如果您错过了月/省，您仍然可以使用dcast：

# Province cases year month #1 Newyork 10 2000 1 #2 Newyork 20 2000 2 #3 Newyork 30 2000 3 #4 Newyork 40 2000 4 #5 LosAngeles 30 2000 1 #6 LosAngeles 40 2000 2 #7 LosAngeles 50 2000 3 #8 LosAngeles 60 2000 4 #9 Newyork 99 2000 5 #10 SanDiego 99 2000 5 dcast(df, Province~month+year, value.var='cases') # Province 1_2000 2_2000 3_2000 4_2000 5_2000 #1 LosAngeles 30 40 50 60 NA #2 Newyork 10 20 30 40 99 #3 SanDiego NA NA NA NA 99

Answer 2

加入“＃month”后，我们可以reshape使用base R。和＆＃39;年＆＃39;列（paste(...)）

 reshape(
    transform(df1, yearmonth=paste('at', month, year, sep="."))[,-(3:4)], 
       idvar='Province', timevar='yearmonth', direction='wide')
#  Province cases.at.1.2000 cases.at.2.2000 cases.at.3.2000    cases.at.4.2000
# 1    Newyork              10              20              30              40
# 5 Los Angeles             30              40              50              60

数据

df1 <- structure(list(Province = c("Newyork", "Newyork", "Newyork", 
"Newyork", "Los Angeles", "Los Angeles", "Los Angeles", "Los Angeles"
), cases = c(10L, 20L, 30L, 40L, 30L, 40L, 50L, 60L), year = c(2000L, 
2000L, 2000L, 2000L, 2000L, 2000L, 2000L, 2000L), month = c(1L, 
 2L, 3L, 4L, 1L, 2L, 3L, 4L)), .Names = c("Province", "cases", 
"year", "month"), class = "data.frame", row.names = c(NA, -8L))

Answer 3

基于@Ananda Mahto的建议：

library(tidyr); library(dplyr)

df %>% mutate(month = paste0("cases.at.", month)) %>%  
  unite(key, month, year, sep=".") %>% spread(key, cases)

如果某个省缺少月 - 年，请使用展开：

df %>% expand(Province, year, month) %>% left_join(df) %>% 
  mutate(month = paste0("cases.at.", month)) %>%  
  unite(key, month, year, sep=".") %>% spread(key, cases)

数据：

df=structure(list(Province = c("Newyork", "Newyork", "Newyork", "Newyork", "LosAngeles", "LosAngeles", "LosAngeles", "LosAngeles", "SanDiego"), cases = c(10L, 20L, 30L, 40L, 30L, 40L, 50L, 60L, 90L), year = c(2000L, 2000L, 2000L, 2000L, 2000L, 2000L, 2000L, 2000L, 2000L), month = c(1L, 2L, 3L, 4L, 1L, 2L, 3L, 4L, 4L)), .Names = c("Province", "cases", "year", "month"), class = "data.frame", row.names = c(NA, -9L))

重塑数据框从长格式到宽格式

3 个答案:

数据