按子段排序数据框

Question

我和我的团队正在处理成千上万个具有相似段的 URL。 某些 URL 在我们感兴趣的位置有一个段（“seg”、复数、“segs”）。其他类似的 URL 在我们感兴趣的位置上有不同的段。 我们需要对由 URL 和相关的唯一段组成的数据框进行排序 在感兴趣的位置，显示这些独特段的频率。

这是一个简化的例子：

 url <- c(1, 3, 1, 4, 2, 3, 1, 3, 3, 3, 3, 2)
 seg <- c("a", "c", "a", "d", "b", "c", "a", "x", "x", "y", "c", "b")
 df <- data.frame(url,seg)

我们正在寻找以下内容：

url freq seg 
 1   3    a   in other words, url #1 appears three times each with a seg = "a",
 2   2    b   in other words: url #2 appears twice each with a seg = "b",
 3   3    c   in other words: url #3 appears three times with a seg = "c", 
 3   2    x                                  two times with a seg = "x", and, 
 3   1    y                                  once with a seg = "y"
 4   1    d   etc.

我可以使用循环和几个小步骤到达那里，但我相信有一种更优雅的方式来做到这一点。这是我的不雅方法：

创建具有 num.unique 行和三列（url、freq、seg）的空数据框

 result <- data.frame(url=0, Freq=0, seg=0)

确定唯一的 URL

 unique.df.url <- unique(df$url)

遍历数据框

 for (xx in unique.df.url) {
   url.seg <- df[which(df$url == unique.df.url[xx]), ] # create a dataframe for each of the unique urls and associated segs
   freq.df.url <- data.frame(table(url.seg))  # summarize the frequency distribution of the segs by url
   result <- rbind(result,freq.df.url)  # append a new data.frame onto the last one
 }

消除数据框中频率 = 0 的行

 result.freq <- result[which(result$Freq |0), ]

按 URL 对数据框进行排序

 result.order <- result.freq[order(result.freq$url), ]

这产生了预期的结果，但由于它太不优雅了，我担心一旦我们扩大规模，所需的时间将会令人望而却步，或者至少是一个问题。有什么建议？

Answer 1

在基础 R 中，您可以这样做：

aggregate(freq~seg+url,`$<-`(df,freq,1),sum)
# or aggregate(freq~seg+url, data.frame(df,freq=1),sum)

#   seg url freq
# 1   a   1    3
# 2   b   2    2
# 3   c   3    3
# 4   x   3    2
# 5   y   3    1
# 6   d   4    1

$<- 的诀窍是在任何地方添加一个值为 1 的列 freq，而不更改您的源表。

另一种可能性：

subset(as.data.frame(table(df[2:1])),Freq!=0)
#    seg url Freq
# 1    a   1    3
# 8    b   2    2
# 15   c   3    3
# 17   x   3    2
# 18   y   3    1
# 22   d   4    1

这里我使用[2:1] 来切换列的顺序，以便table 以所需的方式对结果进行排序。

Answer 2

一个选项可以是使用table 和tidyr::gather 来获取OP 所需格式的数据：

library(tidyverse)
table(df) %>% as.data.frame() %>% 
  filter(Freq > 0 ) %>%
  arrange(url, desc(Freq))


#   url seg  Freq
# 1   1   a     3
# 2   2   b     2
# 3   3   c     3
# 4   3   x     2
# 5   3   y     1
# 6   4   d     1

或

df %>% group_by(url, seg) %>%
  summarise(freq = n()) %>%
  arrange(url, desc(freq))

# # A tibble: 6 x 3
# # Groups: url [4]
#    url seg      freq
#   <dbl> <fctr> <int>
# 1  1.00 a          3
# 2  2.00 b          2
# 3  3.00 c          3
# 4  3.00 x          2
# 5  3.00 y          1
# 6  4.00 d          1

Answer 3

或粘贴并点击：

url <- c(1, 3, 1, 4, 2, 3, 1, 3, 3, 3, 3, 2)
seg <- c("a", "c", "a", "d", "b", "c", "a", "x", "x", "y", "c", "b")
df <- data.frame(url,seg)

want <- tapply(url, INDEX = paste(url, seg, sep = "_"), length)
want <- data.frame(do.call(rbind, strsplit(names(want), "_")), want)
colnames(want) <- c("url", "seg", "freq")
want <- want[order(want$url, -want$freq), ]
rownames(want) <- NULL # needed?
want <- want[ , c("url", "freq", "seg")] # needed?
want

Answer 4

下面的代码对你来说会更好吗？

library(dplyr)
df %>% group_by(url, seg) %>% summarise(n()) 

Answer 5

url <- c(1, 3, 1, 4, 2, 3, 1, 3, 3, 3, 3, 2)
seg <- c("a", "c", "a", "d", "b", "c", "a", "x", "x", "y", "c", "b")
df <- data.frame(url,seg)

library(dplyr)

df %>% count(url, seg) %>% arrange(url, desc(n))

# # A tibble: 6 x 3
#     url seg       n
#   <dbl> <fct> <int>
# 1     1 a         3
# 2     2 b         2
# 3     3 c         3
# 4     3 x         2
# 5     3 y         1
# 6     4 d         1