Question

我的df如下：

mov

我想做两件事。首先，计算具有苹果和橙子（即2玛丽和约翰）的独特观察的数量。

之后，我想将它们从我的数据框中删除，这样我只留下了只有苹果的独特个体。

这是我尝试过的

data
   names  fruit
7   john  apple
13  john orange
14  john  apple
2   mary orange
5   mary  apple
8   mary orange
10  mary  apple
12  mary  apple
1    tom  apple
6    tom  apple

真的，我想使用grepl，因为我的真实数据比水果更复杂。这就是我尝试过的（首先转换为data.table）

toremove<-unique(data[data$fruit=='apple' & data$fruit=='orange',"names"])  ##this part doesn't work, if it had I would have used the below code to remove the names identified
data2<-data[!data$names %in% toremove,]

所以，总而言之，我的问题在于识别同时拥有苹果和橙子的人。这看起来很简单，所以请随意指导我一个可以教我这个的资源！

所需的输出

data1<-data.table(data1)
z<-data1[,ind := grepl('app.*? & orang.*?', fruit), by='names']  ## this works fine when i just use 'app.*?' but collapses when I try to add the & sign, so I'm making an error with the operator. In addition the by='names' doesn't work out for me, which is important. My plan here was to create an indicator (if an individual has an apple and an orange, then they get an indicator==1 and I would then filter them out on the basis of this indicator).

Answer 1

如果您只查找仅包含apple的名称，则此处采用简单的data.table方法

setDT(data)[ , if(all(fruit == "apple")) .SD, by = names]
#    names fruit
# 1:   tom apple
# 2:   tom apple

对于同时具有＆＃34; apple＆＃34;和＆＃34;橙＆＃34;伯爵，你可以做点什么

data[, any(fruit == "apple") & any(fruit == "orange"), by = names][, sum(V1)]
## [1] 2

最后，如果您只想找到只有一个唯一fruit的用户，则可以尝试使用devel version on GH（或uniqueN）<{1}}中的length(unique())进行调节/ p>

data[, if(uniqueN(fruit) < 2L) .SD, by = names]
#    names fruit
# 1:   tom apple
# 2:   tom apple

Answer 2

我使用dplyr包来标记/发现使用橙子的用户和使用这两种水果的用户。（我在最后添加了一行以获得仅有橙色的案例。）

data = 
  read.table(text="
             names  fruit
             7   john  apple
             13  john orange
             14  john  apple
             2   mary orange
             5   mary  apple
             8   mary orange
             10  mary  apple
             12  mary  apple
             1    tom  apple
             6    tom  apple
             21  kathy orange", header=T)

#    names  fruit
# 7   john  apple
# 13  john orange
# 14  john  apple
# 2   mary orange
# 5   mary  apple
# 8   mary orange
# 10  mary  apple
# 12  mary  apple
# 1    tom  apple
# 6    tom  apple
# 21 kathy orange


library(dplyr)

data %>%
  group_by(names) %>%                            # for each user name
  mutate(N_dist = n_distinct(fruit),             # count distinct number of fruits
         N_oranges = sum(fruit=="orange")) %>%   # count number of oranges
  filter(N_oranges == 0 & N_dist < 2) %>%        # keep users with no oranges and no both fruits
  select(names, fruit)


#   names fruit
# 1   tom apple
# 2   tom apple

请注意，在应用过滤器之前，您的数据集如下所示：

#    names  fruit N_dist N_oranges
# 1   john  apple      2         1
# 2   john orange      2         1
# 3   john  apple      2         1
# 4   mary orange      2         2
# 5   mary  apple      2         2
# 6   mary orange      2         2
# 7   mary  apple      2         2
# 8   mary  apple      2         2
# 9    tom  apple      1         0
# 10   tom  apple      1         0
# 11 kathy orange      1         1

您可以从中获得具有水果的唯一名称或具有橙子的用户。

确定满足两个条件的独特观测值，然后去除R.

2 个答案: