当您有具有多个数据的组时，在 R 中计算组标准偏差

Question

我正在使用 R，我正在尝试正确计算我的标准差。

我的数据如下所示：

Target     category       wordproduced       wordValue
wall         A                  home           .003
wall         A                  table          .005
widnow       A                  cow            .015
window       B                  backyard       .012
friend       B                  dog            .018
friend       B                  chance         .088
friend       B                  spoon          .002
big          C                  country        .009
big          C                  pen            .015
big          C                  pub            .012
money        C                  palace         .078
rail         C                  wood           .026
rail         C                  ferrari        .030
rail         C                  car            .062
science      D                  phone          .007
science      D                  laboratory     .009
science      D                  side           .019
water        D                  ocean          .013
water        D                  river          .020

所以，我有四个类别（A、B、C、D），我总共有 8 个单词。每个词都属于一个类别。

所以，如果我想计算目标单词产生的单词的平均值，我会编写这样的代码 ....

mydata %>%
 group_by(category) %>%
 summarise(TargetN = length(unique(Taregt)), 
           wPoroducedN = length(wordsproduced),
           meanW = wProducedN/TargetN)

如果我使用 mean() 函数计算平均值，它会得到错误的平均值，因为它会计算目标中的每个单词。例如，类别 A 只有 2 个唯一词，但总共有 3 个。所以，我需要计算我的平均跳水量为 2。上面的代码解决了这个问题。但是在计算 SD 时，我得到了很多错误的答案或 NA。

例如，我试过这个...

mydata %>%
 group_by(category) %>%
 summarise(TargetN = length(unique(Taregt)), 
           wPoroducedN = length(wordsproduced),
           meanW = wProducedN/TargetN, 
           SD = sd(length(wordproduced)))

在这里，我得到 NA.，而在其他代码中，我得到 0 o o 唯一目标的确切数量，等等。

我应该如何计算我的 SD？

添加可重现的数据.... **类别已更改为数字；相反，os ABCD 是 123（只有三个）

newDat <- structure(list(Target = c(
  "permit",
  "confusion",
  "presion",
  "transanction",
  "sorprise",
  "same",
  "agony",
  "prime",
  "suffer",
  "affect",
  "car",
  "neglect",
  "intern",
  "explore",
  "image",
  "pension",
  "amature",
  "terrified",
  "importance",
  "deal",
  "replace",
  "euforic",
  "optimist",
  "return",
  "inmerse",
  "doll",
  "actor",
  "singular",
  "desctruction",
  "dispute",
  "tremor",
  "profesional",
  "redem",
  "euforic",
  "pen",
  "pause",
  "cultive",
  "center",
  "cheer",
  "slace",
  "recess",
  "apple",
  "introduction",
  "despicable",
  "offense",
  "inteligent",
  "hope",
  "contender",
  "stress",
  "disgust"
), Category = c(
  "3",
  "1",
  "1",
  "1",
  "1",
  "1",
  "1",
  "2",
  "2",
  "2",
  "2",
  "2",
  "1",
  "1",
  "2",
  "2",
  "1",
  "1",
  "2",
  "1",
  "1",
  "1",
  "1",
  "2",
  "1",
  "1",
  "3",
  "1",
  "1",
  "1",
  "1",
  "1",
  "1",
  "1",
  "2",
  "3",
  "1",
  "3",
  "1",
  "2",
  "2",
  "1",
  "1",
  "1",
  "1",
  "2",
  "1",
  "3",
  "1",
  "1"
), wordproduced = c(
  "liberty",
  "intense",
  "sad",
  "serenity",
  "afraid",
  "sadness",
  "hurt",
  "freedom",
  "depress",
  "feeling",
  "love",
  "positive",
  "river",
  "palace",
  "ilusion",
  "stress",
  "aliviated",
  "violence",
  "presion",
  "damage",
  "hate",
  "happy",
  "dwindle",
  "spoon",
  "kitchen",
  "dog",
  "backyard",
  "alone",
  "cat",
  "confidence",
  "fear",
  "moving",
  "house",
  "ocean",
  "territory",
  "continent",
  "sky",
  "rainbow",
  "approach",
  "law",
  "good",
  "school",
  "science",
  "land",
  "laboratory",
  "engage",
  "destiny",
  "voice",
  "arange",
  "infertile"
), wordValue = c(
  0.10,
  0.09,
  0.01,
  0.1,
  0.046,
  0.316,
  0.12,
  0.03,
  0.03,
  0.02,
  0.46,
  0.19,
  0.26,
  0.070,
  0.040,
  0.01,
  0.025,
  0.03,
  0.05,
  0.089,
  0.075,
  0.03,
  0.067,
  0.04,
  0.04,
  0.1,
  0.068,
  0.055,
  0.17,
  0.075,
  0.535,
  0.06,
  0.1,
  0.12,
  0.04,
  0.08,
  0.036,
  0.1,
  0.05,
  0.050,
  0.07,
  0.05,
  0.8,
  0.05,
  0.06,
  0.08,
  0.055,
  0.04,
  0.12,
  0.049
)), row.names = c(NA, -50L), class = c("tbl_df",
                                       "tbl", "data.frame"))

Answer 1

编辑添加的示例数据：

虽然我不确定您要做什么，但我可以告诉您在那里得到NAs，因为您要求的是一个数字的 SD...这没有意义。即...length(wordsproduced) 会给你一个长度的数字，一次一个类别。

我假设您想要每个 target 的 wordsproduced 数量的 SD，对于每个 category。

因此，您计算了 wordsproduced 和 Target 和 category 的平均值，如下所示：

newDat_summary <- newDat %>%
  group_by(Category) %>%
  summarise(TargetN = length(unique(Target)), 
            wProducedN = length(wordproduced),
            meanW = wProducedN/TargetN)

> newDat_summary
# A tibble: 3 x 4
  Category TargetN wProducedN meanW
  <chr>      <int>      <int> <dbl>
1 1             31         32  1.03
2 2             13         13  1   
3 3              5          5  1

对于SD，我们需要先在每个category中分别找到wordsproduced每个Target的个数：

newDat_summary2 <- newDat %>%
  group_by(Category, Target) %>%
  summarise(TargetN = length(unique(Target)), 
            wProducedN = length(wordproduced))

> newDat_summary2
# A tibble: 49 x 4
# Groups:   Category [3]
   Category Target       TargetN wProducedN
   <chr>    <chr>          <int>      <int>
 1 1        agony              1          1
 2 1        amature            1          1
 3 1        apple              1          1
 4 1        cheer              1          1
 5 1        confusion          1          1
 6 1        cultive            1          1
 7 1        deal               1          1
 8 1        desctruction       1          1
 9 1        despicable         1          1
10 1        disgust            1          1
# ... with 39 more rows

现在我们有多个值，我们可以找到它们之间的 SD：

newDat_summary3 <- newDat_summary2 %>% group_by(Category) %>%
  summarise(SD = sd(wProducedN))

> mydata_summary3
# A tibble: 4 x 2
  category    SD
  <chr>    <dbl>
1 A        0.707
2 B        1.41 
3 C        1.15 
4 D        0.707

然后我们通过Target 和category 的方式加入：

newDat_summary <- merge(newDat_summary,newDat_summary3,by = "Category")

> newDat_summary
  Category TargetN wProducedN    meanW        SD
1        1      31         32 1.032258 0.1796053
2        2      13         13 1.000000 0.0000000
3        3       5          5 1.000000 0.0000000

我希望这就是你要找的。​​p>