R代码：错误（函数：参数暗示不同的行数：1、2

Question

以下部分代码对我来说很好用。 Assessments_jobSmart 是一个导入的 csv 文件。我正在尝试重塑多列的数据，但我无法通过使用 tapply 来实现。所以我尝试通过匹配user_id来绑定矩阵，并添加id不存在的行id。

individual_assess <- filter(assessments_jobSmart, !is.na(assessments_jobSmart$submitted))
quiz_i = filter(individual_assess, assessment_type == 'quiz')
checkin_i = filter(individual_assess, assessment_type == 'checkin')


groupQuiz_i <- group_by(quiz_i, user_id, program_id, name)
summaryQuiz_i <- summarize(groupQuiz_i, maxQuiz = max(moderated_score))
q1 <- with(summaryQuiz_i, tapply(maxQuiz, list(user_id, name) , I))
q2=matrix(NA, nrow = nrow(q1), ncol = ncol(q1)+1)
colnames(q2) = c("user_id", paste("assess_q", colnames(q1)))
q2[1:nrow(q2),1] = rownames(q1)
q2[1:nrow(q2),2:ncol(q2)] = q1[1:nrow(q1),1:ncol(q1)]
lla = merge(lla, q2, by = 'user_id', all=TRUE)

但是对于下一部分，它在最后一行的末尾给了我错误。我参考了很多链接，但仍然不知道为什么。 c1 和 c2 的行数相同。

Error in (function (..., row.names = NULL, check.rows = FALSE, check.names = 
TRUE, : arguments imply differing number of rows: 1, 2

错误代码：

groupCheckin_i <- group_by(checkin_i, user_id, program_id, name)
summaryCheckin_i <- summarize(groupCheckin_i, countCheckin = n())
c1 <- with(summaryCheckin_i, tapply(countCheckin, list(user_id, name) , I))
c1[c1=="NULL"]=NA
c2=matrix(NA, nrow = nrow(c1), ncol = ncol(c1)+1)
colnames(c2) = c("user_id", paste("assess_c", colnames(c1)))
c2[1:nrow(c2),1] = rownames(c1)
c2[1:nrow(c2),2:ncol(c2)] = c1[1:nrow(c1),1:ncol(c1)]
lla = merge(lla, c2, by = 'user_id', all=TRUE)

可重复的示例，不确定我是否正确复制它。我将从导入、分组和汇总的数据框开始。

install.packages("dplyr")
install.packages("reshape2")
install.packages('ggplot2', dep = TRUE) 
library("dplyr")
library(reshape2)
library(ggplot2)

head = c("user_id", "program", "assessment", "type", "marks")
content = c("111", "program A", "quiz 1", "quiz", "1", "112", "program A", "quiz 1", "quiz", "0.5", "112", "program A", "quiz 2", "quiz", "0.75", "113", "program B", "quiz 2", "quiz", "0.8", "110", "program B", "survey 1", "survey", "1", "113", "program B", "survey 1", "survey", "1")
M = as.dataframe(matrix(content, nrow=5, ncol=5)) #kinda replicate my imported csv file.
s = filter(M, type == 'survey')
q = filter(M, type == 'quiz')
groupS = group_by(s, user_id, program, assessment)
groupQ = group_by(q, user_id, program, assessment)

summaryS <- summarize(groupS, maxMarks = max(marks)) # take only maximum marks if there are duplicate entries
s1 <- with(summaryS, tapply(maxMarks, list(user_id, assessment) , I))
s2 = matrix(NA, nrow=nrow(s1), ncol=ncol(s1)+1)
colnames(s2) = c("user_id", colnames(s1))
s2[q:nrow(s2), 1] = rownames(s1)      # everything works alright till here
s2[1:nrow(s2),2:ncol(s2)] = s1[1:nrow(s1),1:nrow(s1)]

summaryQ <- summarize(groupQ, count = n()) # it makes more sense to count survey done
q1 <- with(summaryQ, tapply(count, list(user_id, assessment) , I))
q2 = matrix(NA, nrow=nrow(q1), ncol=ncol(q1)+1)
colnames(q2) = c("user_id", colnames(q1))
q2[q:nrow(q2), 1] = rownames(q1)      # everything works alright till here
q2[1:nrow(q2),2:ncol(q2)] = q1[1:nrow(q1),1:nrow(q1)]  #q2 becomes a list :(
b = merge(b, a2, by = 'user_id', all+TRUE)


#     | program   | assessment   | marks
# ---------------------------------------
# 111 | program A | quiz 1       | 1
# 112 | program A | quiz 1       | 0.5
# 112 | program A | quiz 2       | 0.75
# 113 | program B | quiz 2       | 0.8

# Then I used tapply to reshape data to get something like:

#     | assessment 1 | assessment 2
# -------------------------------------
# 111 | 1            | NA
# 112 | 0.5          | 0.75
# 113 | NA           | 0.8

这样的表有很多，因为提取出来的结果不同，所以想在最后合并一下，把所有的结果合并起来。我不希望 user_id 出现两次，每次用于单独的表。我想将结果与 user_id 进行比较，但缺少该列的标题，因为 user_id 被视为行名。所以我创建了一个更大的矩阵来复制所有内容并包含 user_id 列名：

# Let's say a1 is the matrix after tapply, a2 is the new dataframe I want to create, b is the successful new dataframe created using the exact same method on same csv file exported.

我想在合并后得到这样的东西：

user_id | survey 1 | survey 2 | assessment 1 | assessment 2
-------------------------------------------------------------
110     | 1        | NA       | NA           | NA
111     | NA       | NA       | 1            | NA
112     | NA       | NA       | 0.5          | 0.75
113     | 1        | NA       | NA           | 0.8

Answer 1

我仍然很难准确地理解你的出发点和你想要什么。并且您的可重现示例的某些部分不起作用。我试图填补空白，所以这是我试图回答的问题。

您可能想尝试使用tidyr 函数来重塑数据帧（gather/spread 和unite/separate），而不是所有的应用函数。

library(dplyr)
library(tidyr)
library(stringr)

df1 <- data_frame(
           V1 = c("111", "program A", "quiz 1", "quiz", "1"),
           V2 = c("112", "program A", "quiz 1", "quiz", "0.5"),
           V3 = c("112", "program A", "quiz 2", "quiz", "0.75"),
           V4 = c("113", "program B", "quiz 2", "quiz", "0.8"),
           V5 = c("110", "program B", "survey 1", "survey", "1")
)

df1
#> # A tibble: 5 x 5
#>          V1        V2        V3        V4        V5
#>       <chr>     <chr>     <chr>     <chr>     <chr>
#> 1       111       112       112       113       110
#> 2 program A program A program A program B program B
#> 3    quiz 1    quiz 1    quiz 2    quiz 2  survey 1
#> 4      quiz      quiz      quiz      quiz    survey
#> 5         1       0.5      0.75       0.8         1

col_names <- c("user_id", "program", "assessment", "type", "marks")

df2 <- df1 %>% 
  summarise_all(paste, collapse = ",") %>%
  gather("key", "val") %>% 
  separate(val, col_names, ",", convert = TRUE) %>% 
  mutate(assessment = assessment %>% str_extract("\\d") %>% as.integer(),
         program = str_extract(program, ".{1}$")) %>% 
  select(-key)

df2
#> # A tibble: 5 x 5
#>   user_id program assessment   type marks
#>     <int>   <chr>      <int>  <chr> <dbl>
#> 1     111       A          1   quiz  1.00
#> 2     112       A          1   quiz  0.50
#> 3     112       A          2   quiz  0.75
#> 4     113       B          2   quiz  0.80
#> 5     110       B          1 survey  1.00

df3 <- df2 %>% 
  group_by(user_id, program, type, assessment) %>% 
  summarise(marks = max(marks)) %>% 
  unite(key, program, type, assessment) %>% 
  spread(key, marks)

df3
#> Source: local data frame [4 x 5]
#> Groups: user_id [4]
#> 
#> # A tibble: 4 x 5
#>   user_id A_quiz_1 A_quiz_2 B_quiz_2 B_survey_1
#> *   <int>    <dbl>    <dbl>    <dbl>      <dbl>
#> 1     110       NA       NA       NA          1
#> 2     111      1.0       NA       NA         NA
#> 3     112      0.5     0.75       NA         NA
#> 4     113       NA       NA      0.8         NA

您还提到了 user_id 作为某些数据的行名时遇到的问题。 tibble::rownames_to_column() 对此很有帮助。

library(tibble)

head(mtcars)
#>                    mpg cyl disp  hp drat    wt  qsec vs am gear carb
#> Mazda RX4         21.0   6  160 110 3.90 2.620 16.46  0  1    4    4
#> Mazda RX4 Wag     21.0   6  160 110 3.90 2.875 17.02  0  1    4    4
#> Datsun 710        22.8   4  108  93 3.85 2.320 18.61  1  1    4    1
#> Hornet 4 Drive    21.4   6  258 110 3.08 3.215 19.44  1  0    3    1
#> Hornet Sportabout 18.7   8  360 175 3.15 3.440 17.02  0  0    3    2
#> Valiant           18.1   6  225 105 2.76 3.460 20.22  1  0    3    1

mtcars %>% 
  rownames_to_column("model") %>% 
  head()
#>               model  mpg cyl disp  hp drat    wt  qsec vs am gear carb
#> 1         Mazda RX4 21.0   6  160 110 3.90 2.620 16.46  0  1    4    4
#> 2     Mazda RX4 Wag 21.0   6  160 110 3.90 2.875 17.02  0  1    4    4
#> 3        Datsun 710 22.8   4  108  93 3.85 2.320 18.61  1  1    4    1
#> 4    Hornet 4 Drive 21.4   6  258 110 3.08 3.215 19.44  1  0    3    1
#> 5 Hornet Sportabout 18.7   8  360 175 3.15 3.440 17.02  0  0    3    2
#> 6           Valiant 18.1   6  225 105 2.76 3.460 20.22  1  0    3    1

此外，由于您使用了很多 tidyverse 函数，因此您可能需要查看 dplyr::*_join 函数而不是合并。