Question

ODK响应

1
1; 2
1; 2; 3
1; 2; 3; 5
1; 2; 4
1; 2; 4; 5; 6
1; 2; 4; 6
1; 2; 4; 7

1 is Crop failure-
2 is Water shortage
3 is Land degradation
4 is Lack of HH Labor
5 is Lack of income from agriculture
6 is Lack of manure / fertilizer
7 is Others

我想要一张这样的桌子

Crop failure- 8
Water shortage- 7
Land degradation- 6
Lack of HH Labor- 1
Lack of income from agriculture- 2
Lack of manure / fertilizer- 2
Others- 1

我曾在R中使用dplyr在“将具有多个值的单列拆分为多个列”中尝试过，但请帮忙。

Answer 1

使用{'1': 240, '2': 60}，您可以获得以下信息：

plyr

您会得到：

Condition = c("Crop failure", "Water shortage", "Lang degradation", "Lack of HH Labor",  "Lack of income from agriculture", "Lack of manure / fertilizer", "Others")
Type = c(1:7)
df = data.frame(Condition, Type)

vector = c(1,1,2,1,2,3,1,2,3,5,1,2,4,1,2,4,5,6,1,2,4,6,1,2,4,7)
t = plyr::count(vector)
colnames(t) = c("Type","Freq")

df =merge(df,t)

Answer 2

如果您想使用base R，以下解决方案可能会对您有所帮助。

假设您输入的是

response <- c(1, 1, 2, 1, 2, 3, 1, 2, 3, 5, 1, 2, 4, 1, 2, 4, 5, 6, 1, 2, 4, 6, 1, 2, 4, 7)

然后

status <- c("Crop failure", "Water shortage", "Lang degradation", "Lack of HH Labor",  "Lack of income from agriculture", "Lack of manure / fertilizer", "Others")
df <- as.data.frame(table(factor(response,labels = status),dnn = list("Status")))

可以给您类似的输出

> df
                           Status Freq
1                    Crop failure    8
2                  Water shortage    7
3                Lang degradation    2
4                Lack of HH Labor    4
5 Lack of income from agriculture    2
6     Lack of manure / fertilizer    2
7                          Others    1

如果您想要详细的表格：假设您输入的是：

r <- list(1, c(1, 2), c(1, 2, 3), c(1, 2, 3, 5), c(1, 2, 4), c(1, 
2, 4, 5, 6), c(1, 2, 4), 6, c(1, 2, 4, 7))

type = seq(1,7)
dt <- as.data.frame(t(sapply(r, function(v) sapply(type, function(k) sum(k==v)))))
colnames(M) <- paste0("type",type)

给出

> dt
  type1 type2 type3 type4 type5 type6 type7
1     1     0     0     0     0     0     0
2     1     1     0     0     0     0     0
3     1     1     1     0     0     0     0
4     1     1     1     0     1     0     0
5     1     1     0     1     0     0     0
6     1     1     0     1     1     1     0
7     1     1     0     1     0     0     0
8     0     0     0     0     0     1     0
9     1     1     0     1     0     0     1

此外，每个类型条目的总和可以通过colSums计算：

> colSums(dt)
type1 type2 type3 type4 type5 type6 type7 
    8     7     2     4     2     2     1

或者您可以使用match()，即

dt <- as.data.frame(t(sapply(r, function(v) !is.na(match(type,v)))))
> dt
  type1 type2 type3 type4 type5 type6 type7
1  TRUE FALSE FALSE FALSE FALSE FALSE FALSE
2  TRUE  TRUE FALSE FALSE FALSE FALSE FALSE
3  TRUE  TRUE  TRUE FALSE FALSE FALSE FALSE
4  TRUE  TRUE  TRUE FALSE  TRUE FALSE FALSE
5  TRUE  TRUE FALSE  TRUE FALSE FALSE FALSE
6  TRUE  TRUE FALSE  TRUE  TRUE  TRUE FALSE
7  TRUE  TRUE FALSE  TRUE FALSE FALSE FALSE
8 FALSE FALSE FALSE FALSE FALSE  TRUE FALSE
9  TRUE  TRUE FALSE  TRUE FALSE FALSE  TRUE

Answer 3

这是我解决问题的机会。我使用tidyverse，因为它为我加载了stringer和tidyr

library(tidyverse)

id <- data.frame(Code = 1:7, #Make a coding data frame so you can label the results
                 Cause = c("Crop failure", "Water shortage", "Land degradation", "Lack of HH Labor", "Lack of income from agriculture", "Lack of manure / fertilizer", "Others"), stringsAsFactors = FALSE))

data <- Book1 %>% 
  separate(X1, into = paste0("X", 1:7), sep = ";") %>% #split the data by the ;, This induces NA that are removed later
  gather(key = "drop", value = "Code") %>% #put it into 1 column to exploit R's vectorization
  mutate(Code = as.integer(Code)) %>% #Make the code an integer for the join later
  filter(!is.na(Code)) %>% #remove those previous NAs
  group_by(Code) %>% 
  count() %>% # Counts 
  left_join(., id) #labels

colnames(data) <- c("Code", "Count", "Cause")

它将在单独的行上发出警告，但是，这只是让您知道它正在用NA填充多余的单元格，我们稍后将其删除。您唯一需要更改的就是DataFrame和X1，具体取决于您为对象命名的名称。

这是我的搜索结果

 Code Count Cause                          
  <int> <int> <chr>                          
1     1     8 Crop failure                   
2     2     7 Water shortage                 
3     3     2 Land degradation               
4     4     4 Lack of HH Labor               
5     5     2 Lack of income from agriculture
6     6     2 Lack of manure / fertilizer    
7     7     1 Others

希望有帮助！

如何将列分成多个列并在R中找到频率？

3 个答案: