将每一行除以其总和

Question

我有这个数据：

 merged_dat1
# A tibble: 4 x 35
# Groups:   Product.Name [4]
  Product.Name  also  apps battery better   big camera  case cheap   day definitely enough  even however
  <chr>        <int> <int>   <int>  <int> <int>  <int> <int> <int> <int>      <int>  <int> <int>   <int>
1 BLU Studio ~   498   545    1552    465   306    839   406   161   413        225    156   590     178
2 iphone 4s       71    37     380     43    19     13    21    38    43         25     16   128      52
3 Motorola Mo~   160   221     253    123    69    428    39   125   101         49    157   133      79
4 Samsung Gal~    76    70     122     49    25     69    22    17    31         15     53    71      31
# ... with 21 more variables: issues <int>, life <int>, little <int>, long <int>, lot <int>, low <int>,
#   many <int>, memory <int>, much <int>, overall <int>, phones <int>, pictures <int>, pretty <int>,
#   quality <int>, right <int>, screen <int>, size <int>, still <int>, use <int>, way <int>, well <int>

我想将一行的每个值除以它的行总和。

例如，第一行的行和为15044 。 第一行的每个值都应该除以这个值。 这可能是重复的，但我找不到关于此的主题。 如果有人有dplyr解决方案，那就太好了。

structure(list(Product.Name = c("BLU Studio 5.0", "iphone 4s", 
"Motorola Moto E", "Samsung Galaxy II"), also = c(498L, 71L, 
160L, 76L), apps = c(545L, 37L, 221L, 70L), battery = c(1552L, 
380L, 253L, 122L), better = c(465L, 43L, 123L, 49L), big = c(306L, 
19L, 69L, 25L), camera = c(839L, 13L, 428L, 69L), case = c(406L, 
21L, 39L, 22L), cheap = c(161L, 38L, 125L, 17L), day = c(413L, 
43L, 101L, 31L), definitely = c(225L, 25L, 49L, 15L), enough = c(156L, 
16L, 157L, 53L), even = c(590L, 128L, 133L, 71L), however = c(178L, 
52L, 79L, 31L), issues = c(334L, 49L, 60L, 23L), life = c(649L, 
60L, 136L, 25L), little = c(283L, 45L, 156L, 44L), long = c(197L, 
49L, 65L, 25L), lot = c(316L, 35L, 107L, 39L), low = c(203L, 
25L, 116L, 24L), many = c(207L, 32L, 77L, 51L), memory = c(200L, 
10L, 148L, 48L), much = c(421L, 79L, 165L, 53L), overall = c(206L, 
35L, 77L, 8L), phones = c(749L, 84L, 214L, 63L), pictures = c(263L, 
12L, 94L, 32L), pretty = c(332L, 25L, 97L, 31L), quality = c(669L, 
40L, 186L, 49L), right = c(189L, 49L, 45L, 33L), screen = c(1359L, 
71L, 252L, 82L), size = c(244L, 7L, 93L, 55L), still = c(416L, 
48L, 107L, 28L), use = c(650L, 126L, 256L, 140L), way = c(218L, 
40L, 44L, 12L), well = c(605L, 103L, 205L, 114L)), class = c("grouped_df", 
"tbl_df", "tbl", "data.frame"), row.names = c(NA, -4L), vars = "Product.Name", drop = TRUE, indices = list(
    0L, 1L, 2L, 3L), group_sizes = c(1L, 1L, 1L, 1L), biggest_group_size = 1L, labels = structure(list(
    Product.Name = c("BLU Studio 5.0", "iphone 4s", "Motorola Moto E", 
    "Samsung Galaxy II")), class = "data.frame", row.names = c(NA, 
-4L), vars = "Product.Name", drop = TRUE))

Answer 1

我们可以先创建一个rowSums列，然后再做mutate_if

library(tidyverse)
merged_dat1 %>% 
   as.data.frame %>%
   mutate(newSum = select_if(., is.numeric) %>% 
                       reduce(`+`)) %>% 
   mutate_if(is.numeric, list(~ ./newSum)) %>% 
   select(-newSum)
#  Product.Name       also       apps    battery     better         big      camera       case      cheap
#1    BLU Studio 5.0 0.03310290 0.03622707 0.10316405 0.03090933 0.020340335 0.055769742 0.02698750 0.01070194
#2         iphone 4s 0.03717277 0.01937173 0.19895288 0.02251309 0.009947644 0.006806283 0.01099476 0.01989529
#3   Motorola Moto E 0.03450507 0.04766013 0.05456114 0.02652577 0.014880311 0.092301057 0.00841061 0.02695708
#4 Samsung Galaxy II 0.04662577 0.04294479 0.07484663 0.03006135 0.015337423 0.042331288 0.01349693 0.01042945
#         day  definitely      enough       even    however     issues       life     little       long        lot
#1 0.02745281 0.014956129 0.010369583 0.03921829 0.01183196 0.02220154 0.04314012 0.01881149 0.01309492 0.02100505
#2 0.02251309 0.013089005 0.008376963 0.06701571 0.02722513 0.02565445 0.03141361 0.02356021 0.02565445 0.01832461
#3 0.02178132 0.010567177 0.033858098 0.02868234 0.01703688 0.01293940 0.02932931 0.03364244 0.01401768 0.02307526
#4 0.01901840 0.009202454 0.032515337 0.04355828 0.01901840 0.01411043 0.01533742 0.02699387 0.01533742 0.02392638
#         low       many      memory       much     overall     phones    pictures     pretty    quality      right
#1 0.01349375 0.01375964 0.013294337 0.02798458 0.013693167 0.04978729 0.017482053 0.02206860 0.04446956 0.01256315
#2 0.01308901 0.01675393 0.005235602 0.04136126 0.018324607 0.04397906 0.006282723 0.01308901 0.02094241 0.02565445
#3 0.02501617 0.01660556 0.031917188 0.03558335 0.016605564 0.04615053 0.020271727 0.02091870 0.04011214 0.00970455
#4 0.01472393 0.03128834 0.029447853 0.03251534 0.004907975 0.03865031 0.019631902 0.01901840 0.03006135 0.02024540
#      screen        size      still        use         way       well
#1 0.09033502 0.016219091 0.02765222 0.04320659 0.014490827 0.04021537
#2 0.03717277 0.003664921 0.02513089 0.06596859 0.020942408 0.05392670
#3 0.05434548 0.020056071 0.02307526 0.05520811 0.009488894 0.04420962
#4 0.05030675 0.033742331 0.01717791 0.08588957 0.007361963 0.06993865

注意：这也会进行类型检查

---

此外，如果我们base R ，这可以紧凑地完成为

merged_dat1[-1] <- merged_dat1[-1]/rowSums(merged_dat1[-1])

Answer 2

使用dplyr ，您还可以尝试：

df %>%
 ungroup() %>%
 mutate(across(-1)/rowSums(across(-1)))

  Product.Name   also   apps battery better     big  camera    case  cheap    day
  <chr>         <dbl>  <dbl>   <dbl>  <dbl>   <dbl>   <dbl>   <dbl>  <dbl>  <dbl>
1 BLU Studio … 0.0331 0.0362  0.103  0.0309 0.0203  0.0558  0.0270  0.0107 0.0275
2 iphone 4s    0.0372 0.0194  0.199  0.0225 0.00995 0.00681 0.0110  0.0199 0.0225
3 Motorola Mo… 0.0345 0.0477  0.0546 0.0265 0.0149  0.0923  0.00841 0.0270 0.0218
4 Samsung Gal… 0.0466 0.0429  0.0748 0.0301 0.0153  0.0423  0.0135  0.0104 0.0190

或按列类型求和：

df %>%
 ungroup() %>%
 mutate(across(where(is.numeric))/rowSums(across(where(is.numeric))))