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首页 > 解决方案 > R代码：向现有数据添加一列（平均分数）。帧不返回正确的数字

r - R代码：向现有数据添加一列（平均分数）。帧不返回正确的数字

问题描述

我正在尝试清理数据集，以便轻松分析它。

我有一个如下所示的数据集：

              z  a   b   c   d  a_1   b_2   c_3   d_4  ab_1  ab_2
Participant1  A  1   3   4   3   2     3     6     7     2    1
Participant2  B  3   2   4   5   6     4     2     1     2    1
Participant3  C  1   3   5   4   2     2     1     2     3    4
Participant4  D  1   3   8   1   3     4     5     4     5    2

我为每个测试变量添加了一列平均分数。因此，我使用了以下代码：

CleanData <- CleanData%>%
  group_by(ResponseId) %>%
  mutate(MeanA = mean(as.numeric(c(a, b, c, d)))) %>%
  mutate(MeanB = mean(as.numeric(c(a_1, b_2, c_3, d_4)))) %>%
  mutate(MeanC = mean(as.numeric(c(ab_1, ab_2))))

使用上述代码，我能够向现有数据框添加三个新列（MeanA、MeanB 和 MeanC），即：

            Z a   b   c   d  a_1   b_2   c_3   d_4  ab_1  ab_2 meanA meanB meancC
Participant1 A 1   3   4   3   2     3     6     7     2    1   2.75   4,5    1,5
Participant2 B 3   2   4   5   6     4     2     1     2    1     x     x     x
Participant3 C 1   3   5   4   2     2     1     2     3    4     x     x     x
Participant4 D 1   3   8   1   3     4     5     4     5    2     x     x     x

但是，平均列不显示正确的数字。例如，值 ab_1: 3 和 ab_2: 4 的平均值应返回 3.5，但该列显示 5。谁能帮我理解为什么会这样？

谢谢

更新：我发现值：a_1、b_2、c_3、d_4、ab_1、ab_2 是因子，所以我通过添加将它们更改为数字，例如：

as.numeric(as.character(mean(c(ab_1, ab_2))))

但它仍然没有返回正确的值！！

UPDATE2： dput 返回以下

structure(list(ResponseId = structure(c(32L, 40L, 113L, 43L), .Label = c("R_10UwSkTd0ZY0I6s", 
"R_12z514ODiHL80r6", "R_1Bx5NSYrdqhaIuJ", "R_1C8bt9iRS6L1817", 
"R_1C91Jv4yYTt8VsS", "R_1Cj5l50lu2VKcxg", "R_1DD2zbJPiueDijH", 
"R_1FA36tDxZEUqBJM", "R_1FLH4KQaKlxMn77", "R_1FeUYLmPoDpKnOb", 
"R_1GAXJ8YmAJflif4", "R_1GE4YeIfEuZuUYX", "R_1GyaiQcync3sHMJ", 
"R_1IRUh7uRVEI3Op1", "R_1KjzzPz48WkuCFk", "R_1LcddBAuXLiBDgo", 
"R_1M65lFIT9h0TrpQ", "R_1NCw01UV09WSGSX", "R_1QlN6pk1IH5xXqn", 
"R_1XMp4A95WZLJmWl", "R_1dBn0Bb9ZPprg3R", "R_1dcu8FxLC7E2tq8", 
"R_1dgBt0IGuRjDD4W", "R_1eRGDucOhbsOkgI", "R_1eXovzU5BJzWdt9", 
"R_1g5Unx7VkpjcHEe", "R_1gtnPQhBie1xBLq", "R_1guMjae0AF4k9yh", 
"R_1j6xFfGDt6QCgc3", "R_1jOf096UI4ManbG", "R_1l3smK23l8TPmso", 
"R_1ohW9lO1Qw7D2Nt", "R_1pDO3wCW2J7X9Mw", "R_1qU6JvhOWUPen4Y", 
"R_23e2rcSv4MzoilL", "R_27W3bSNMVuVHxLl", "R_2Bqy2lJu3vKgTkC", 
"R_2PiG0aapXGpeAhi", "R_2PjUnRQM4nSaYuO", "R_2QnhIx6nPDMKt4J", 
"R_2S1saV61VMsVhzA", "R_2SjpkptLr59Pi5W", "R_2VQn7jIGZe7rSiJ", 
"R_2VgwMZugDIzh2gc", "R_2Y3ylZL7Os57FQg", "R_2YtPLlRDm3W0IiF", 
"R_2cjhW6bLhpKoWeD", "R_2czLWk4DA7AtUWB", "R_2dhg48SvjsPF7xi", 
"R_2e3xcJj4B4sVxpu", "R_2fcI8PF5YakOibH", "R_2qqi75IiKtupT3k", 
"R_2s705LLk7dRpBtu", "R_2tKAG3kO0uTAp3W", "R_2tglcBV7ZIr5fRL", 
"R_2uWFRjzrewJ9lZp", "R_2w6pFIySy6z417N", "R_2wRKqbJfulfII8L", 
"R_2zTr8igVCFw8pBZ", "R_2zbYrfdxoSMQDZw", "R_2zqoRmevtHcELGs", 
"R_32XQhbZdKS1gZqP", "R_33C0jdizQOZOCam", "R_3CWxNxWuLPFvTzi", 
"R_3DbBdhc5Z5gYa8p", "R_3ESZKQIuMZpxjWb", "R_3HhHnbzqwkjBTyr", 
"R_3L76M1sxyZT8LZF", "R_3LecRRD5DH4kAHZ", "R_3NwVV3CjA62MD4l", 
"R_3Oj2rtphCHA2eid", "R_3OjKwwD8cFQtLqs", "R_3e3rPFQew2oBcZB", 
"R_3gT9FFFMBZ0iVhF", "R_3h6MPewodtUKKEi", "R_3lAoiHlF9KgfbFB", 
"R_3lFbGqMCvw5n4bo", "R_3nAFNfIRZmI8wzE", "R_3nTXeACPBHRGkZI", 
"R_3q7MZwtck0MNgI0", "R_3qwfz0yMLuWw81M", "R_3suKETbTurYGy0m", 
"R_3yleFr647Wb8EQV", "R_55a2UXtA2wqJeXD", "R_5w2RvbtXFaQpLMd", 
"R_6Gs5YBPSE2abloR", "R_6LFgmHhwol4jAEp", "R_9WRZaEXexVKfN4t", 
"R_AMNieClr3z90yuB", "R_BtXxoHRS0zenny1", "R_Cg1tqfa6iBeV6Ct", 
"R_DiBGGojPdv9dDjz", "R_DjKGN63D9nlkRoJ", "R_DzUYyku3vvPVHgd", 
"R_OlJDf3crqx5ixC9", "R_T6pA7HhF1QdSffz", "R_UlL0oXl78e1DAWJ", 
"R_Wk3USlMJv0smEXD", "R_XRO3jxTdfN1ATyV", "R_XyLTjQJ24ZC2WnT", 
"R_Z3o0VAV64k7sUmt", "R_ZltqETejsEhSOE9", "R_b4xwkSOfYOui3nP", 
"R_bPCma1oTtGdWKpr", "R_cuAayrJroDlIlep", "R_dclKNvKcXlAfHs5", 
"R_dgr9ooE75LVOpWh", "R_e2oMPa2WBgFmW9H", "R_eF0zW2RqPXuWtah", 
"R_eKfj3XyRPgRlEsh", "R_oXC0xDrC6uq8uEp", "R_qPowYCxUMjc2Bz3", 
"R_sMtw9IU4BTmSBMt", "R_tVxiAfQAU1TpQ8p", "R_vq3xb7ppDytKW4x", 
"R_x8TnBj3YBj2R345", "R_xitfzNOT9yXK1vr", "R_yJSFhJqUec5dHfH", 
"Response ID", "{\"ImportId\":\"_recordId\"}"), class = "factor"), 
    FL_8_DO = structure(c(6L, 6L, 1L, 1L), .Label = c("", "FL_27", 
    "FL_28", "FL_8 - Block Randomizer - Display Order", "F", 
    "H", "{\"ImportId\":\"FL_8_DO\"}"), class = "factor"), 
    Gender = structure(c(2L, 2L, 1L, 2L), .Label = c("", "1", 
    "2", "what is your gender?", "{\"ImportId\":\"QID14\"}"), class = "factor"), 
    Age = structure(c(2L, 10L, 1L, 7L), .Label = c("", "10", 
    "12", "14", "20", "3", "4", "5", "6", "7", "8", "9", "how old are you?", 
    "{\"ImportId\":\"QID27\"}"), class = "factor"), Per_1 = c(4, 
    2, NA, 1), Per_2 = c(3, 2, NA, 1), A_1 = c(NA_real_, 
    NA_real_, NA_real_, NA_real_), A_2 = c(NA_real_, NA_real_, 
    NA_real_, NA_real_), A_3 = c(NA_real_, NA_real_, NA_real_, 
    NA_real_), A_4 = c(NA_real_, NA_real_, NA_real_, NA_real_
    ), A_5 = c(NA_real_, NA_real_, NA_real_, NA_real_), A_6 = c(NA_real_, 
    NA_real_, NA_real_, NA_real_), A_7 = c(NA_real_, NA_real_, 
    NA_real_, NA_real_), A_8 = c(NA_real_, NA_real_, NA_real_, 
    NA_real_), A_9 = c(NA_real_, NA_real_, NA_real_, NA_real_
    ), Mo_1 = c(NA_real_, NA_real_, NA_real_, NA_real_), 
   Mo_2 = c(NA_real_, NA_real_, NA_real_, NA_real_), Mo_3 = c(NA_real_, 
    NA_real_, NA_real_, NA_real_), Mo_4 = c(NA_real_, NA_real_, 
    NA_real_, NA_real_), MeanA = c(NA_real_, NA_real_, NA_real_, 
    NA_real_)), .Names = c("ResponseId", "FL_8_DO", "Gender", 
"Age", "Per_1", "Per_2", "A_1", "A_2", "A_3", 
"A_4", "A_5", "A_6", "A_7", "A_8", "A_9", "Mo_1", 
"Mo_2", "Mo_3", "Mo_4", "MeanA"), row.names = c(NA, 
4L), class = "data.frame")

标签： rdataframedplyr

解决方案

嗨，问题是mutate对列进行操作。

考虑这个小例子：

df <- tibble(a = 1:3, b = 4:6)
df %>% mutate(x = mean(c(a, b)))
    # A tibble: 3 x 3
      a     b     x
  <int> <int> <dbl>
1     1     4   3.5
2     2     5   3.5
3     3     6   3.5

mutate 所做的是将它堆叠在一起a，b从而得到向量 1,2,3,4,5,6，然后计算其平均值，即 3.5。

为了mutate逐行操作，您必须使用以下rowwise功能：

df %>% rowwise() %>%  mutate(x = mean(c(a, b)))
# A tibble: 3 x 3
      a     b     x
  <int> <int> <dbl>
1     1     4   2.5
2     2     5   3.5
3     3     6   4.5

更新：如果您有属于类的列，factor则应首先将它们转换为数字。如本问题所述，有几种方法可以做到这一点。

在您的情况下，您可以使用

df %>% mutate_at(vars("a", "b", "c", "d"), ~as.numeric(levels(.x))[.x])

将 a、b、c 和 d 列从因子转换为数值。

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