Question

我的桌子看起来像这样：

 df <- read.table(text = 
      "  Day            location     gender    hashtags
       'Feb 19 2016'       'UK'      'M'       '#a'
       'Feb 19 2016'       'UK'      'M'       '#b'
       'Feb 19 2016'       'SP'      'F'       '#a'
       'Feb 19 2016'       'SP'      'F'       '#b'
       'Feb 19 2016'       'SP'      'M'       '#a'
       'Feb 19 2016'       'SP'      'M'       '#b'
       'Feb 20 2016'       'UK'      'F'       '#a'", 
                 header = TRUE, stringsAsFactors = FALSE)

我想按天/标签/位置和性别计算频率，结果表如下所示：

           Day hashtags Daily_Freq men women Freq_UK Freq_SP
   Feb 19 2016       #a          3   2     1       1       2
   Feb 19 2016       #b          3   2     1       1       1
   Feb 20 2016       #a          1   0     1       1       0

其中Daily_freq =男性+女性= Freq_UK + Freq_SP 我怎么能这样做？

Answer 1

使用dplyr：

library(dplyr)
df %>% 
  group_by(Day, hashtags) %>% 
  summarise(Daily_Freq = n(),
            men = sum(gender == 'M'),
            women = sum(gender == 'F'),
            Freq_UK = sum(location == 'UK'),
            Freq_SP = sum(location == 'SP'))

给出：

# A tibble: 3 x 7
# Groups:   Day [?]
  Day         hashtags Daily_Freq   men women Freq_UK Freq_SP
  <chr>       <chr>         <int> <int> <int>   <int>   <int>
1 Feb 19 2016 #a                3     2     1       1       2
2 Feb 19 2016 #b                3     2     1       1       2
3 Feb 20 2016 #a                1     0     1       1       0

data.table中实现的逻辑相同：

library(data.table)
setDT(df)[, .(Daily_Freq = .N,
              men = sum(gender == 'M'),
              women = sum(gender == 'F'),
              Freq_UK = sum(location == 'UK'),
              Freq_SP = sum(location == 'SP'))
          , by = .(Day, hashtags)]

Answer 2

单程......

library(data.table)

setDT(df)
df[, gender := as.factor(gender)]
df[, location := as.factor(location)]

df[, c(
  N = .N, 
  dcast(.SD, . ~ gender, fun.agg = length, drop=FALSE)[, !"."],
  dcast(.SD, . ~ location, fun.agg = length, drop=FALSE)[, !"."]
), by=.(Day, hashtags)]

#            Day hashtags N F M SP UK
# 1: Feb 19 2016       #a 3 1 2  2  1
# 2: Feb 19 2016       #b 3 1 2  2  1
# 3: Feb 20 2016       #a 1 1 0  0  1

以这种方式编码可能更容易维护：不需要手动分配列名;地点和性别将根据它们是否出现在原始数据中而显示或退出结果;和列名称不需要在多个位置输入（转换为因子后）。

如果国家/地区代码与性别代码匹配，则这种方式会产生重复的列。绕过那个：

df[, c(
  N = .N, 
  gender = dcast(.SD, . ~ gender, fun.agg = length, drop=FALSE)[, !"."],
  loc = dcast(.SD, . ~ location, fun.agg = length, drop=FALSE)[, !"."]
), by=.(Day, hashtags)]

#            Day hashtags N gender.F gender.M loc.SP loc.UK
# 1: Feb 19 2016       #a 3        1        2      2      1
# 2: Feb 19 2016       #b 3        1        2      2      1
# 3: Feb 20 2016       #a 1        1        0      0      1

Answer 3

使用包reshape2。

library(reshape2)

molten <- melt(df, id.vars = c("Day", "hashtags"))
result <- dcast(molten, Day + hashtags ~ variable + value, length)
result$Daily_Freq <- rowSums(result[, c("location_SP", "location_UK")])

result
#          Day hashtags location_SP location_UK gender_F gender_M Daily_Freq
#1 Feb 19 2016       #a           2           1        1        2          3
#2 Feb 19 2016       #b           2           1        1        2          3
#3 Feb 20 2016       #a           0           1        1        0          1

请注意，列不是示例输出的顺序。重新排序它们很简单。

按某些列汇总单元格

3 个答案: