计算每个因子按另一个因子分组的数量

Question

我知道这个问题的答案很简单，但是我已经在各个论坛中进行了广泛搜索，但一直找不到解决方案。

我有一列名为Data_source的列，这是我要将变量分组的一个因素。

我有一系列symptom*变量，我想要根据Data_source进行计数。

由于某种原因，我无法弄清楚该怎么做。普通的group_by函数似乎无法正常工作。

这是有问题的数据框

 df <- wrapr::build_frame(
   "Data_source"  , "Sex"   , "symptoms_decLOC", "symptoms_nausea_vomitting" |
     "1"          , "Female", NA_character_    , NA_character_               |
     "1"          , "Female", NA_character_    , NA_character_               |
     "1"          , "Female", "No"             , NA_character_               |
     "1"          , "Female", "Yes"            , "No"                        |
     "1"          , "Female", "Yes"            , "No"                        |
     "1"          , "Female", "Yes"            , "No"                        |
     "1"          , "Male"  , "Yes"            , "No"                        |
     "1"          , "Female", "Yes"            , "No"                        |
     "2"          , "Female", NA_character_    , NA_character_               |
     "2"          , "Male"  , NA_character_    , NA_character_               |
     "2"          , "Male"  , NA_character_    , NA_character_               |
     "2"          , "Female", "Yes"            , "No"                        |
     "2"          , "Female", "Yes"            , "No"                        |
     "2"          , "Male"  , NA_character_    , NA_character_               |
     "2"          , "Male"  , NA_character_    , NA_character_               |
     "2"          , "Male"  , NA_character_    , NA_character_               |
     "2"          , "Female", NA_character_    , NA_character_               |
     "2"          , "Female", NA_character_    , NA_character_               |
     "2"          , "Male"  , NA_character_    , NA_character_               |
     "2"          , "Female", NA_character_    , NA_character_               )

请注意，性别和症状变量都是包括NA在内的所有因素。我尝试了以下

df %>% na.omit() %>% group_by(Data_source) %>% count("symptoms_decLOC")

这行不通，并且不是最佳选择，因为我必须对每一列都重复一次。理想的做法是使用类似于lapply(df, count)的名称，但这并不能给我每个组的描述。

编辑

为回答以下问题，我添加了预期的输出。我已经在excel中对此进行了编辑，为清楚起见，对group_by进行了颜色编码。

请注意，我如何对每个可能的答案进行细分。当我使用dplyr运行此命令时，输出如下。

> df %>% na.omit() %>% group_by(Data_source) %>% count("symptoms_decLOC")
# A tibble: 2 x 3
# Groups:   Data_source [2]
  Data_source `"symptoms_decLOC"`     n
  <chr>       <chr>               <int>
1 1           symptoms_decLOC         5
2 2           symptoms_decLOC         2

Answer 1

这是最有效的方法：尚未弄清如何包括零计数组...应该添加.drop=FALSE takes care of this，但对我不起作用（使用std::vector v。0.8 .0.9001）。

dplyr

结果：

library(dplyr)
library(tidyr)
(df
    %>% tidyr::gather(var,val,-Data_source)
    %>% count(Data_source,var,val, .drop=FALSE)
    %>% na.omit()
)

Answer 2

使用@Ben Bolker的答案来获取每个组的计数，并使用spread和gather包括零计数组。

dplyr

library(dplyr)
library(tidyr)

# Count number of occurences by Data_source 
df2 <- 
  df %>% 
  gather(variable, value, -Data_source) %>% 
  count(Data_source, variable, value, name = "counter") %>%
  na.omit() 

# For variable = "Sex", leave as is
# For everything else, in this case symptom* convert into factor to include zero count group
# Then spread with dataframe will NAs filled with 0, re-convert back to long to bind rows
bind_rows(df2 %>%
            filter(variable == "Sex"), 

          df2 %>%
            filter(variable != "Sex") %>%
            mutate(value = factor(value, levels = c("Yes", "No"))) %>%
            spread(key = value, value = counter, fill = 0) %>%
            gather(value, counter, -Data_source, -variable))  %>%

  arrange(Data_source, variable)

data.table

library(data.table)
dt <- data.table(df)

# Melt data by Data source
dt_melt <- melt(dt, id.vars = "Data_source", value.factor = FALSE, variable.factor = FALSE)

# Add counter, if NA then 0 else 1
dt_melt[, counter := 0]
dt_melt[!is.na(value), counter := 1]

# Sum number of occurrences
dt_count <- dt_melt[,list(counter = sum(counter)), by = c("Data_source", "variable", "value")]

# Split into two dt
dt2a <- dt_count[variable == "Sex", ]
dt2b <- dt_count[variable != "Sex" ,]

# only on symptoms variables
# Convert into factor variable
dt2b$value <- factor(dt2b$value, levels = c("Yes", "No"))
dt2b_dcast <- dcast(data = dt2b, formula = Data_source + variable ~ value, value.var = "counter", fill = 0, drop = FALSE)
dt2b_melt <- melt(dt2b_dcast, id.vars = c("Data_source", "variable"), variable.name = "value", value.name = "counter") 

# combine
combined_d <- rbind(dt2a, dt2b_melt)
combined_d[order(Data_source, variable), ]

Answer 3

我不太了解您的要求，但我假设您要计算每个symptom_*列中非NA值的数量。

这是一个data.table解决方案：

# load library

library(data.table)

# Suppose the table is called "dt". Convert it to a data.table:

setDT(dt)

# convert the wide table to a long one, filter the values that
# aren't NA and count both, by Data_source and by variable
# (variable is the created column with the symptom_* names)

melt(dt, id.vars = 1:2)[!is.na(value), 
                        .N, 
                         by = .(Data_source, variable)]

代码的每个部分在做什么：

melt(dt, id.vars = 1:2)将dt从宽转换为长，并将第1列和第2列（Data_source和sex）保持固定。

!is.na(value)过滤不是symptom_*的值（以前在每个NA标题下）。

.N对行进行计数。

by = .(Data_source, variable)是我们用来计算的分组。 variable是symptom_*在整形期间到达的列的名称。

Answer 4

绝对，困难的是保持数据中不存在的组合...这是分两个步骤的解决方案：

1。准备没有计数的数据库

您可以做任何您想做的事，但是我选择计算两个块，因为变量Sex的方式不同。无需在此处绑定这些块。

chunk1 <- expand.grid(
  Data_source = c("1", "2"),
  name = c("symptoms_decLOC", "symptoms_nausea_vomitting"),
  value = c("Yes", "No"),
  stringsAsFactors = FALSE
)

chunk2 <- expand.grid(
  Data_source = c("1", "2"),
  name = "Sex",
  value = c("Female", "Male"),
  stringsAsFactors = FALSE
)

2。完成要求的工作

library(dplyr)
library(tidyr)

df %>% 
  pivot_longer(cols = c("Sex", "symptoms_decLOC", "symptoms_nausea_vomitting"))%>%
  group_by(Data_source, name, value) %>%
  summarise(count = n()) %>%
  right_join(bind_rows(chunk1, chunk2), by = c("Data_source", "name", "value")) %>%
  arrange(Data_source, name) %>%
  mutate(count = zoo::na.fill(count, 0))

Etvoilà

# A tibble: 12 x 4
# Groups:   Data_source, name [6]
   Data_source name                      value  count
   <chr>       <chr>                     <chr>  <int>
 1 1           Sex                       Female     7
 2 1           Sex                       Male       1
 3 1           symptoms_decLOC           Yes        5
 4 1           symptoms_decLOC           No         1
 5 1           symptoms_nausea_vomitting Yes        0
 6 1           symptoms_nausea_vomitting No         5
 7 2           Sex                       Female     6
 8 2           Sex                       Male       6
 9 2           symptoms_decLOC           Yes        2
10 2           symptoms_decLOC           No         0
11 2           symptoms_nausea_vomitting Yes        0
12 2           symptoms_nausea_vomitting No         2

它不是那么短，但是它使用简单的功能。该过程类似于在Excel中可以完成的过程，即准备结构，然后完成计数。

我希望它可以帮助;-）