2

我有一个长数据格式的数据集：

Date        Region     X   Y   Z   T    D  E   F 
01-01-2020  RegionA    2   4   2   3   2   3   4
01-01-2020  RegionB    1   3   2   2   3   3   3
01-01-2020  RegionC    1   4   4   2   3   4   2
01-01-2020  RegionD    2   4   2   3   2   4   4
01-01-2020  RegionE    1   3   2   2   2   2   2
02-01-2020  RegionA    2   4   7   3   2   3   4
02-01-2020  RegionB    1   3   2   2   2   3   3
02-01-2020  RegionC    1   4   4   8   3   4   2
02-01-2020  RegionD    2   3   2   3   2   4   4
02-01-2020  RegionE    1   3   2   2   2   2   2

日期还有很多，但这应该让您对格式有所了解。

然后我有第二个数据集，其中包含有关这些地区人口的更多信息：

Region     Pop
RegionA    2000
RegionB    4039
RegionC    24728  
RegionD    3738  
RegionE    2936

我想要做的是将第一个数据集中的一列除以每个地区的人口值，跨越所有日期。例如，如果“x”是GDP我想除以GDP每个不同时间点的总体值。对于RegionA这将是2/2000和2/2000对于每个01-01-2020和02-01-2020。

我对 R 很陌生，任何帮助开始解决这个问题都会很棒。

这里有一个可重现的例子

date<-as.Date(c("2020-02-24T18:00:00", "2020-02-24T18:00:00", "2020-02- 
                 24T18:00:00", "2020-05-02T17:00:00", "2020-05-02T17:00:00", 
                 "2020-05-02T17:00:00"))
regions<-c("RegionA", "RegionB", "RegionC","RegionA", "RegionB", "RegionC")
total<-c(1394, 1143, 18373, 168479, 65370, 26990)
df<-data.frame(date, regions, total)

对于另一个数据框：

regions<-c("RegionA", "RegionB", "RegionC")
pop<-c(1305283, 559084, 1935414)
mydf_pop<-data.frame(regions, pop)

现在：我尝试了各种组合

df >%>
  left_join(mydf_pop)>%>
  group_by(date, regions)>%>
  mutate(total/pop)

这显然是错误的。

谢谢你。

4

2 回答 2

1

您可以使用一个left_join和mutate：

library(dplyr)

new_df <- left_join(df1, df2, by = "Region")

new_df %>%
   mutate_at(vars(X:F), ~ . / Pop)

我们通过 common 列将两个数据集连接在一起Region，然后通过重新mutate_at计算变量。创建一个公式对象，点指的是被引用的列。XF~.

这给了我们：

# A tibble: 10 x 10
   Date       Region          X        Y        Z         T        D        E         F   Pop
   <chr>      <chr>       <dbl>    <dbl>    <dbl>     <dbl>    <dbl>    <dbl>     <dbl> <dbl>
 1 01-01-2020 RegionA 0.001     0.002    0.001    0.0015    0.001    0.0015   0.002      2000
 2 01-01-2020 RegionB 0.000248  0.000743 0.000495 0.000495  0.000743 0.000743 0.000743   4039
 3 01-01-2020 RegionC 0.0000404 0.000162 0.000162 0.0000809 0.000121 0.000162 0.0000809 24728
 4 01-01-2020 RegionD 0.000535  0.00107  0.000535 0.000803  0.000535 0.00107  0.00107    3738
 5 01-01-2020 RegionE 0.000341  0.00102  0.000681 0.000681  0.000681 0.000681 0.000681   2936
 6 02-01-2020 RegionA 0.001     0.002    0.0035   0.0015    0.001    0.0015   0.002      2000
 7 02-01-2020 RegionB 0.000248  0.000743 0.000495 0.000495  0.000495 0.000743 0.000743   4039
 8 02-01-2020 RegionC 0.0000404 0.000162 0.000162 0.000324  0.000121 0.000162 0.0000809 24728
 9 02-01-2020 RegionD 0.000535  0.000803 0.000535 0.000803  0.000535 0.00107  0.00107    3738
10 02-01-2020 RegionE 0.000341  0.00102  0.000681 0.000681  0.000681 0.000681 0.000681   2936
于 2020-05-14T13:52:33.160 回答
1

一种基本 R 选项是使用match

df1[-c(1:2)] <- df1[-(1:2)]/df2$Pop[match(df1$Region,df2$Region)]

这使

> df1
         Date  Region            X            Y            Z            T
1  01-01-2020 RegionA 1.000000e-03 0.0020000000 0.0010000000 1.500000e-03
2  01-01-2020 RegionB 2.475860e-04 0.0007427581 0.0004951721 4.951721e-04
3  01-01-2020 RegionC 4.043999e-05 0.0001617599 0.0001617599 8.087997e-05
4  01-01-2020 RegionD 5.350455e-04 0.0010700910 0.0005350455 8.025682e-04
5  01-01-2020 RegionE 3.405995e-04 0.0010217984 0.0006811989 6.811989e-04
6  02-01-2020 RegionA 1.000000e-03 0.0020000000 0.0035000000 1.500000e-03
7  02-01-2020 RegionB 2.475860e-04 0.0007427581 0.0004951721 4.951721e-04
8  02-01-2020 RegionC 4.043999e-05 0.0001617599 0.0001617599 3.235199e-04
9  02-01-2020 RegionD 5.350455e-04 0.0008025682 0.0005350455 8.025682e-04
10 02-01-2020 RegionE 3.405995e-04 0.0010217984 0.0006811989 6.811989e-04
              D            E            F
1  0.0010000000 0.0015000000 2.000000e-03
2  0.0007427581 0.0007427581 7.427581e-04
3  0.0001213200 0.0001617599 8.087997e-05
4  0.0005350455 0.0010700910 1.070091e-03
5  0.0006811989 0.0006811989 6.811989e-04
6  0.0010000000 0.0015000000 2.000000e-03
7  0.0004951721 0.0007427581 7.427581e-04
8  0.0001213200 0.0001617599 8.087997e-05
9  0.0005350455 0.0010700910 1.070091e-03
10 0.0006811989 0.0006811989 6.811989e-04

数据

> dput(df1)
structure(list(Date = c("01-01-2020", "01-01-2020", "01-01-2020",
"01-01-2020", "01-01-2020", "02-01-2020", "02-01-2020", "02-01-2020", 
"02-01-2020", "02-01-2020"), Region = c("RegionA", "RegionB",
"RegionC", "RegionD", "RegionE", "RegionA", "RegionB", "RegionC",
"RegionD", "RegionE"), X = c(2L, 1L, 1L, 2L, 1L, 2L, 1L, 1L, 
2L, 1L), Y = c(4L, 3L, 4L, 4L, 3L, 4L, 3L, 4L, 3L, 3L), Z = c(2L,
2L, 4L, 2L, 2L, 7L, 2L, 4L, 2L, 2L), T = c(3L, 2L, 2L, 3L, 2L, 
3L, 2L, 8L, 3L, 2L), D = c(2L, 3L, 3L, 2L, 2L, 2L, 2L, 3L, 2L,
2L), E = c(3L, 3L, 4L, 4L, 2L, 3L, 3L, 4L, 4L, 2L), F = c(4L,
3L, 2L, 4L, 2L, 4L, 3L, 2L, 4L, 2L)), class = "data.frame", row.names = c(NA, 
-10L))

> dput(df2)
structure(list(Region = c("RegionA", "RegionB", "RegionC", "RegionD",
"RegionE"), Pop = c(2000L, 4039L, 24728L, 3738L, 2936L)), class = "data.frame", row.names = c(NA, 
-5L))
于 2020-05-14T14:04:24.383 回答