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将每列的值乘以R中另一个data.frame中的权重

multiplication multiple-columns dataframe r

Sylvia Rodriguez · 技术社区 · 4 年前

我有两个 data.frames df 和 weights

测向 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  stackoverflow   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

重量 data.frame ?

    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  stackoverflow   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", 
      "stackoverflow", "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])

1 回复 | 直到 4 年前

Ronak Shah 4 年前

你可以用 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  stackoverflow  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

TarJae 4 年前

下面是另一种解决方案:

Names 在里面 weights
match 列名来自 df 带列属于 重量
乘以相应的 V1 weights$V1
最后 cbind 得到 id 在里面 测向 ,已被删除 df[-1]

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

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  stackoverflow  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