将每列中的值乘以 R 中另一个 data.frame 中的权重
Multiply values across each column by weight in another data.frame in R
我有两个 data.frames：df 和 weights（代码如下）。
df 看起来像这样：
id a b d EE f
1 this 0.23421153 -0.02324956 0.5457353 0.73068586 0.5642554
2 is 0.28378641 0.36346241 1.0190496 1.97715019 -1.190179
3 an -0.04372133 0.07412557 -0.4510299 1.8442713 -0.5301328
4 example -0.18139887 0.10404329 -1.0871962 1.46920108 0.4685703
5 data.frame 0.24235498 -0.1501064 -1.1183967 0.22884407 0.4272259
6 for -0.72718239 0.16337997 1.2635683 0.44206945 0.7303647
7 Whosebug 0.25203942 -0.1772715 -0.3371532 -0.29167792 -0.7065494
8 please -0.11047364 -0.06631552 0.4342659 -1.49584522 0.2837016
9 help -0.1136639 0.22414253 0.4284864 1.59096047 0.2915938
10 me -0.3677288 0.05974474 -0.1136177 0.02322094 -0.6533994
如何将每列中的值乘以 weights data.frame 中的相应权重？
预期结果：
id a b d EE f
1 this 0.749476896 -0.1743717 5.29363241 NA 4.17548996
2 is 0.908116512 2.725968075 9.88478112 NA -8.8073246
3 an -0.139908256 0.555941775 -4.37499003 NA -3.92298272
4 example -0.580476384 0.780324675 -10.54580314 NA 3.46742022
5 data.frame 0.775535936 -1.125798 -10.84844799 NA 3.16147166
6 for -2.326983648 1.225349775 12.25661251 NA 5.40469878
7 Whosebug 0.806526144 -1.32953625 -3.27038604 NA -5.22846556
8 please -0.353515648 -0.4973664 4.21237923 NA 2.09939184
9 help -0.36372448 1.681068975 4.15631808 NA 2.15779412
10 me -1.17673216 0.44808555 -1.10209169 NA -4.83515556
代码：
set.seed(12345)
df <- data.frame(id=c("this", "is", "an", "example", "data.frame", "for",
"Whosebug", "please", "help", "me"), a=rnorm(10,0,0.4), b=rnorm(10,0,0.2),
d=rnorm(10,0,0.7), EE=rnorm(10,0,0.9), f=rnorm(10,0,0.5))
weights <- data.frame(V1=as.numeric(c("3.2", "7.5", "2.2", "9.7", "5.4", "7.4", "2.1",
"5.0", "3.3", "7.6", "3.6", "7.7", "7.1", "3.3", "9.8", "9.2", "2.5", "6.2", "4.1", "8.7",
"3.3", "9.3", "8.3")))
rownames(weights) <- paste(letters[1:23])
您可以使用 sweep 和 match -
df[-1] <- sweep(df[-1],2, weights$V1[match(names(df[-1]),rownames(weights))],`*`)
df
# id a b d EE f
#1 this 0.7494769 -0.1743717 5.293633 NA 4.175490
#2 is 0.9081165 2.7259681 9.884781 NA -8.807325
#3 an -0.1399082 0.5559418 -4.374990 NA -3.922983
#4 example -0.5804764 0.7803247 -10.545803 NA 3.467420
#5 data.frame 0.7755359 -1.1257980 -10.848448 NA 3.161471
#6 for -2.3269836 1.2253498 12.256612 NA 5.404699
#7 Whosebug 0.8065261 -1.3295363 -3.270386 NA -5.228465
#8 please -0.3535157 -0.4973664 4.212379 NA 2.099392
#9 help -0.3637245 1.6810690 4.156319 NA 2.157794
#10 me -1.1767322 0.4480855 -1.102092 NA -4.835156
这是一个替代解决方案：
- 在
weights 中创建名为 Names 的行名列
match 来自 df 的列名和 weights 数据框的 Names 列 - 通过复制
weights$V1 ，根据数字索引乘以相应的V1
- 最后
cbind 在 df 中得到 id，为了计算目的，它被 df[-1] 删除了。
library(tibble)
# rownames to column
weights <- weights %>%
rownames_to_column(var = "Names")
df_result <- df[-1]*weights$V1[match(names(df[-1]), weights$Names)][col(df[-1])]
df_result <- cbind(id=df$id, df_result)
df_result
输出：
id a b d EE f
1 this 0.7494769 -0.1743717 5.293633 NA 4.175490
2 is 0.9081165 2.7259681 9.884781 NA -8.807325
3 an -0.1399082 0.5559418 -4.374990 NA -3.922983
4 example -0.5804764 0.7803247 -10.545803 NA 3.467420
5 data.frame 0.7755359 -1.1257980 -10.848448 NA 3.161471
6 for -2.3269836 1.2253498 12.256612 NA 5.404699
7 Whosebug 0.8065261 -1.3295363 -3.270386 NA -5.228465
8 please -0.3535157 -0.4973664 4.212379 NA 2.099392
9 help -0.3637245 1.6810690 4.156319 NA 2.157794
10 me -1.1767322 0.4480855 -1.102092 NA -4.835156
使用lapply。应该比 matching 和 sweeping 快得多。
df[-1] <- lapply(names(df)[-1], \(x) df[, x]*weights[x, ])
# id a b d EE f
# 1 this 0.7494769 -0.1743717 5.293633 NA 4.175490
# 2 is 0.9081165 2.7259681 9.884781 NA -8.807325
# 3 an -0.1399082 0.5559418 -4.374990 NA -3.922983
# 4 example -0.5804764 0.7803247 -10.545803 NA 3.467420
# 5 data.frame 0.7755359 -1.1257980 -10.848448 NA 3.161471
# 6 for -2.3269836 1.2253498 12.256612 NA 5.404699
# 7 Whosebug 0.8065261 -1.3295363 -3.270386 NA -5.228465
# 8 please -0.3535157 -0.4973664 4.212379 NA 2.099392
# 9 help -0.3637245 1.6810690 4.156319 NA 2.157794
# 10 me -1.1767322 0.4480855 -1.102092 NA -4.835156
使用 tidyverse，我们可以循环 across 列，获取相应的列名 (cur_column())，使用它通过指定 [=14] 对“权重”数据进行子集化=], 名字和相乘
library(dplyr)
df %>%
mutate(across(where(is.numeric), ~ . * weights[cur_column(), 'V1']))
id a b d EE f
1 this 0.7494769 -0.1743717 5.293633 NA 4.175490
2 is 0.9081165 2.7259681 9.884781 NA -8.807325
3 an -0.1399082 0.5559418 -4.374990 NA -3.922983
4 example -0.5804764 0.7803247 -10.545803 NA 3.467420
5 data.frame 0.7755359 -1.1257980 -10.848448 NA 3.161471
6 for -2.3269836 1.2253498 12.256612 NA 5.404699
7 Whosebug 0.8065261 -1.3295363 -3.270386 NA -5.228465
8 please -0.3535157 -0.4973664 4.212379 NA 2.099392
9 help -0.3637245 1.6810690 4.156319 NA 2.157794
10 me -1.1767322 0.4480855 -1.102092 NA -4.835156
您可以尝试下面的基本 R 代码
df[-1] <- df[-1] * weights[names(df)[-1], "V1"][col(df[-1])]
这给出了
> df
id a b d EE f
1 this 0.7494769 -0.1743717 5.293633 NA 4.175490
2 is 0.9081165 2.7259681 9.884781 NA -8.807325
3 an -0.1399082 0.5559418 -4.374990 NA -3.922983
4 example -0.5804764 0.7803247 -10.545803 NA 3.467420
5 data.frame 0.7755359 -1.1257980 -10.848448 NA 3.161471
6 for -2.3269836 1.2253498 12.256612 NA 5.404699
7 Whosebug 0.8065261 -1.3295363 -3.270386 NA -5.228465
8 please -0.3535157 -0.4973664 4.212379 NA 2.099392
9 help -0.3637245 1.6810690 4.156319 NA 2.157794
10 me -1.1767322 0.4480855 -1.102092 NA -4.835156
我有两个 data.frames：df 和 weights（代码如下）。
df 看起来像这样：
id a b d EE f
1 this 0.23421153 -0.02324956 0.5457353 0.73068586 0.5642554
2 is 0.28378641 0.36346241 1.0190496 1.97715019 -1.190179
3 an -0.04372133 0.07412557 -0.4510299 1.8442713 -0.5301328
4 example -0.18139887 0.10404329 -1.0871962 1.46920108 0.4685703
5 data.frame 0.24235498 -0.1501064 -1.1183967 0.22884407 0.4272259
6 for -0.72718239 0.16337997 1.2635683 0.44206945 0.7303647
7 Whosebug 0.25203942 -0.1772715 -0.3371532 -0.29167792 -0.7065494
8 please -0.11047364 -0.06631552 0.4342659 -1.49584522 0.2837016
9 help -0.1136639 0.22414253 0.4284864 1.59096047 0.2915938
10 me -0.3677288 0.05974474 -0.1136177 0.02322094 -0.6533994
如何将每列中的值乘以 weights data.frame 中的相应权重？
预期结果：
id a b d EE f
1 this 0.749476896 -0.1743717 5.29363241 NA 4.17548996
2 is 0.908116512 2.725968075 9.88478112 NA -8.8073246
3 an -0.139908256 0.555941775 -4.37499003 NA -3.92298272
4 example -0.580476384 0.780324675 -10.54580314 NA 3.46742022
5 data.frame 0.775535936 -1.125798 -10.84844799 NA 3.16147166
6 for -2.326983648 1.225349775 12.25661251 NA 5.40469878
7 Whosebug 0.806526144 -1.32953625 -3.27038604 NA -5.22846556
8 please -0.353515648 -0.4973664 4.21237923 NA 2.09939184
9 help -0.36372448 1.681068975 4.15631808 NA 2.15779412
10 me -1.17673216 0.44808555 -1.10209169 NA -4.83515556
代码：
set.seed(12345)
df <- data.frame(id=c("this", "is", "an", "example", "data.frame", "for",
"Whosebug", "please", "help", "me"), a=rnorm(10,0,0.4), b=rnorm(10,0,0.2),
d=rnorm(10,0,0.7), EE=rnorm(10,0,0.9), f=rnorm(10,0,0.5))
weights <- data.frame(V1=as.numeric(c("3.2", "7.5", "2.2", "9.7", "5.4", "7.4", "2.1",
"5.0", "3.3", "7.6", "3.6", "7.7", "7.1", "3.3", "9.8", "9.2", "2.5", "6.2", "4.1", "8.7",
"3.3", "9.3", "8.3")))
rownames(weights) <- paste(letters[1:23])
您可以使用 sweep 和 match -
df[-1] <- sweep(df[-1],2, weights$V1[match(names(df[-1]),rownames(weights))],`*`)
df
# id a b d EE f
#1 this 0.7494769 -0.1743717 5.293633 NA 4.175490
#2 is 0.9081165 2.7259681 9.884781 NA -8.807325
#3 an -0.1399082 0.5559418 -4.374990 NA -3.922983
#4 example -0.5804764 0.7803247 -10.545803 NA 3.467420
#5 data.frame 0.7755359 -1.1257980 -10.848448 NA 3.161471
#6 for -2.3269836 1.2253498 12.256612 NA 5.404699
#7 Whosebug 0.8065261 -1.3295363 -3.270386 NA -5.228465
#8 please -0.3535157 -0.4973664 4.212379 NA 2.099392
#9 help -0.3637245 1.6810690 4.156319 NA 2.157794
#10 me -1.1767322 0.4480855 -1.102092 NA -4.835156
这是一个替代解决方案：
- 在
weights 中创建名为 match来自df的列名和weights数据框的Names列- 通过复制
weights$V1，根据数字索引乘以相应的 - 最后
cbind在df中得到id，为了计算目的，它被df[-1]删除了。
Names 的行名列
V1
library(tibble)
# rownames to column
weights <- weights %>%
rownames_to_column(var = "Names")
df_result <- df[-1]*weights$V1[match(names(df[-1]), weights$Names)][col(df[-1])]
df_result <- cbind(id=df$id, df_result)
df_result
输出：
id a b d EE f
1 this 0.7494769 -0.1743717 5.293633 NA 4.175490
2 is 0.9081165 2.7259681 9.884781 NA -8.807325
3 an -0.1399082 0.5559418 -4.374990 NA -3.922983
4 example -0.5804764 0.7803247 -10.545803 NA 3.467420
5 data.frame 0.7755359 -1.1257980 -10.848448 NA 3.161471
6 for -2.3269836 1.2253498 12.256612 NA 5.404699
7 Whosebug 0.8065261 -1.3295363 -3.270386 NA -5.228465
8 please -0.3535157 -0.4973664 4.212379 NA 2.099392
9 help -0.3637245 1.6810690 4.156319 NA 2.157794
10 me -1.1767322 0.4480855 -1.102092 NA -4.835156
使用lapply。应该比 matching 和 sweeping 快得多。
df[-1] <- lapply(names(df)[-1], \(x) df[, x]*weights[x, ])
# id a b d EE f
# 1 this 0.7494769 -0.1743717 5.293633 NA 4.175490
# 2 is 0.9081165 2.7259681 9.884781 NA -8.807325
# 3 an -0.1399082 0.5559418 -4.374990 NA -3.922983
# 4 example -0.5804764 0.7803247 -10.545803 NA 3.467420
# 5 data.frame 0.7755359 -1.1257980 -10.848448 NA 3.161471
# 6 for -2.3269836 1.2253498 12.256612 NA 5.404699
# 7 Whosebug 0.8065261 -1.3295363 -3.270386 NA -5.228465
# 8 please -0.3535157 -0.4973664 4.212379 NA 2.099392
# 9 help -0.3637245 1.6810690 4.156319 NA 2.157794
# 10 me -1.1767322 0.4480855 -1.102092 NA -4.835156
使用 tidyverse，我们可以循环 across 列，获取相应的列名 (cur_column())，使用它通过指定 [=14] 对“权重”数据进行子集化=], 名字和相乘
library(dplyr)
df %>%
mutate(across(where(is.numeric), ~ . * weights[cur_column(), 'V1']))
id a b d EE f
1 this 0.7494769 -0.1743717 5.293633 NA 4.175490
2 is 0.9081165 2.7259681 9.884781 NA -8.807325
3 an -0.1399082 0.5559418 -4.374990 NA -3.922983
4 example -0.5804764 0.7803247 -10.545803 NA 3.467420
5 data.frame 0.7755359 -1.1257980 -10.848448 NA 3.161471
6 for -2.3269836 1.2253498 12.256612 NA 5.404699
7 Whosebug 0.8065261 -1.3295363 -3.270386 NA -5.228465
8 please -0.3535157 -0.4973664 4.212379 NA 2.099392
9 help -0.3637245 1.6810690 4.156319 NA 2.157794
10 me -1.1767322 0.4480855 -1.102092 NA -4.835156
您可以尝试下面的基本 R 代码
df[-1] <- df[-1] * weights[names(df)[-1], "V1"][col(df[-1])]
这给出了
> df
id a b d EE f
1 this 0.7494769 -0.1743717 5.293633 NA 4.175490
2 is 0.9081165 2.7259681 9.884781 NA -8.807325
3 an -0.1399082 0.5559418 -4.374990 NA -3.922983
4 example -0.5804764 0.7803247 -10.545803 NA 3.467420
5 data.frame 0.7755359 -1.1257980 -10.848448 NA 3.161471
6 for -2.3269836 1.2253498 12.256612 NA 5.404699
7 Whosebug 0.8065261 -1.3295363 -3.270386 NA -5.228465
8 please -0.3535157 -0.4973664 4.212379 NA 2.099392
9 help -0.3637245 1.6810690 4.156319 NA 2.157794
10 me -1.1767322 0.4480855 -1.102092 NA -4.835156