Question

在过去20年里，我有一些美国 - 朝鲜进口和出口产品，许多列名都是月份的名称，进口月份为IJAN，出口月份为EJAN，因此我使用了gather（）两次尝试以正确的tidydata格式获取它们。

这是我最初的反复：

      # A tibble: 26 x 29
   year  CTY_CODE CTYNAME    IJAN  IFEB  IMAR  IAPR  IMAY  IJUN  IJUL  IAUG  ISEP  IOCT  INOV  IDEC   IYR  EJAN  EFEB  EMAR  EAPR  EMAY  EJUN  EJUL  EAUG  ESEP
   <chr> <chr>    <chr>     <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
 1 1992  5790     Korea, N~    0.    0.    0.    0.    0.    0.    0.    0.    0. 0.       0.    0. 0.     0.   0.100  0.   0.    0.    0.     0.   0.    0.   
 2 1993  5790     Korea, N~    0.    0.    0.    0.    0.    0.    0.    0.    0. 0.       0.    0. 0.     0.   0.     0.   0.    0.    0.     0.   0.    0.   
 3 1994  5790     Korea, N~    0.    0.    0.    0.    0.    0.    0.    0.    0. 0.       0.    0. 0.     0.   0.     0.   0.    0.    0.200  0.   0.    0.   
 4 1995  5790     Korea, N~    0.    0.    0.    0.    0.    0.    0.    0.    0. 0.       0.    0. 0.     0.   0.     6.60 0.    0.    4.20   0.   0.    0.200
 5 1996  5790     Korea, N~    0.    0.    0.    0.    0.    0.    0.    0.    0. 0.       0.    0. 0.     0.   0.     0.   0.400 0.    0.     0.   0.    0.100
 6 1997  5790     Korea, N~    0.    0.    0.    0.    0.    0.    0.    0.    0. 0.       0.    0. 0.     0.   0.     0.   0.    0.100 0.     2.00 0.    0.300
 7 1998  5790     Korea, N~    0.    0.    0.    0.    0.    0.    0.    0.    0. 0.       0.    0. 0.     0.   0.     4.00 0.    0.    0.100  0.   0.300 0.   
 8 1999  5790     Korea, N~    0.    0.    0.    0.    0.    0.    0.    0.    0. 0.       0.    0. 0.     0.   0.100  0.   0.    0.300 0.     1.10 0.500 0.500
 9 2000  5790     Korea, N~    0.    0.    0.    0.    0.    0.    0.    0.    0. 0.100    0.    0. 0.100  2.50 0.100  0.   0.    0.    0.     0.   0.100 0.   
10 2001  5790     Korea, N~    0.    0.    0.    0.    0.    0.    0.    0.    0. 0.       0.    0. 0.     0.   0.     0.   0.    0.    0.     0.   0.    0.   
# ... with 16 more rows, and 4 more variables: EOCT <dbl>, ENOV <dbl>, EDEC <dbl>, EYR <dbl>

我第一次使用gather（）来处理导入月份，它可以正常工作

USNKTrade <- USNKTrade %>% gather(contains("I"), key="month", value="ImportAmount")

导致

# A tibble: 338 x 18
   year  CTY_CODE CTYNAME       EJAN  EFEB  EMAR  EAPR  EMAY  EJUN  EJUL  EAUG  ESEP  EOCT  ENOV  EDEC    EYR month ImportAmount
   <chr> <chr>    <chr>        <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>  <dbl> <chr>        <dbl>
 1 1992  5790     Korea, North  0.   0.100  0.   0.    0.    0.     0.   0.    0.    0.    0.    0.     0.100 IJAN            0.
 2 1993  5790     Korea, North  0.   0.     0.   0.    0.    0.     0.   0.    0.    0.    2.00  0.     2.00  IJAN            0.
 3 1994  5790     Korea, North  0.   0.     0.   0.    0.    0.200  0.   0.    0.    0.    0.    0.     0.200 IJAN            0.
 4 1995  5790     Korea, North  0.   0.     6.60 0.    0.    4.20   0.   0.    0.200 0.    0.500 0.100 11.6   IJAN            0.
 5 1996  5790     Korea, North  0.   0.     0.   0.400 0.    0.     0.   0.    0.100 0.    0.    0.     0.500 IJAN            0.
 6 1997  5790     Korea, North  0.   0.     0.   0.    0.100 0.     2.00 0.    0.300 0.    0.100 0.     2.50  IJAN            0.
 7 1998  5790     Korea, North  0.   0.     4.00 0.    0.    0.100  0.   0.300 0.    0.    0.    0.     4.40  IJAN            0.
 8 1999  5790     Korea, North  0.   0.100  0.   0.    0.300 0.     1.10 0.500 0.500 0.500 0.600 7.70  11.3   IJAN            0.
 9 2000  5790     Korea, North  2.50 0.100  0.   0.    0.    0.     0.   0.100 0.    0.    0.    0.     2.70  IJAN            0.
10 2001  5790     Korea, North  0.   0.     0.   0.    0.    0.     0.   0.    0.    0.    0.    0.500  0.500 IJAN            0.
# ... with 328 more rows

但是当我尝试使用

对导出执行相同操作时

USNKTrade <- USNKTrade %>% gather(starts_with("E"), key="month", value="ExportAmount")

然后我得到一个大约4000行的tibble，实际上我想要一个只有前一个大两倍的tibble（这样每个导入和导出月都有自己的行）。

非常感谢帮助！感谢

Answer 1

这是一种方法。使用starts_with()代替contains()，因为有些月份可能不仅包含E或I的第一个字母。在第二次收集时，请注意我已取消选择新创建的列ImportAmount和ImportMonth以及列year。

year <- 1992:2001
EJAN <- rnorm(10)
IJAN <- rnorm(10)
EFEB <- rnorm(10)
IFEB <- rnorm(10)

df <- data.frame(year, EJAN, IJAN, EFEB, IFEB)
df %>% gather(starts_with("E"), key = ImportMonth, value = ImportAmount) %>%
  gather(starts_with("I"), key = ExportMonth, value = ExportAmount, -ImportAmount, -ImportMonth, - year)

#    year ImportMonth ImportAmount ExportMonth ExportAmount
# 1  1992        EJAN      -1.1528        IJAN      0.94967
# 2  1993        EJAN       0.1165        IJAN      0.86506
# 3  1994        EJAN       0.2553        IJAN     -0.05108
# 4  1995        EJAN      -0.8516        IJAN     -0.50873
# 5  1996        EJAN      -0.3014        IJAN      0.50614
# 6  1997        EJAN       1.4017        IJAN      1.73527
# 7  1998        EJAN       0.8019        IJAN     -0.71507
# 8  1999        EJAN       1.7179        IJAN     -0.32709
# 9  2000        EJAN      -1.2478        IJAN     -1.07364
# 10 2001        EJAN      -1.0491        IJAN     -1.83764
# ...
# 40 2001        EFEB       1.2181        IFEB      0.76118

Answer 2

虽然OP未提供可重现的数据，但似乎问题是OP已递归gather data.frame。

 #The columns containing "I" will move to row after below command
 USNKTrade %>% gather(contains("I"), key="month", value="ImportAmount")

#Now below command will add for more rows
USNKTrade <- USNKTrade %>% gather(starts_with("E"), key="month", value="ExportAmount")

可行的解决方案可能是：

USNKTrade %>% select_at(vars(year, CTY_CODE, CTYNAME, contains("I"))) %>%
           gather(contains("I"), key="month", value="ImportAmount") %>%
inner_join(USNKTrade %>% select_at(vars(year, CTY_CODE, CTYNAME, starts_with("E"))) %>%
              starts_with(contains("E"), key="month", value="ImportAmount"),
         by = c("year", "CTY_CODE", "CTYNAME", "month"))

方法是首先仅选择ImportAmount所需的列，然后应用gather。现在只选择ExportAmount所需的列，然后应用gather。合并两组。

假设加入时"year", "CTY_CODE", "CTYNAME", "month"为key列。

当我使用gather（）两次时，我得到的结果比它应该大得多

2 个答案: