r - 如何在不使用as_tbl_graph（）的情况下创建相同的结果？(how can I create identical result without using as_tbl_graph()?)

Question

My dataframe looks like this.

(我的数据框看起来像这样。)

Word1    Word2    Count
--------------------------
 a         b        4

 c         a        2

 b         c        1
-------------------------

I want the following result.

(我想要以下结果。)

from      to    count
-----------------------
  1       3       4

  2       1       2

  3       2       1

----------------------

I know I can achieve this easily using as_tbl_graph(df).

(我知道我可以使用as_tbl_graph（df）轻松实现这一目标。)

But I want this result only using base r code without using other packages.

(但是我只希望使用base r代码而不使用其他软件包来获得此结果。)

How can I create identical result without using other packages such as igraph, ggraph, tidyverse ...?

(如何在不使用igraph，ggraph，tidyverse等其他软件包的情况下创建相同的结果？)

  ask by jjw translate from so

Answer 1

You can convert the values to factor and then integer to accomplish that:

(您可以将值转换为factor然后再整数以完成此操作：)

lvls <- unique(df$Word1)                    # first we create an object containing the levels found in Word1

df$Word1 <- factor(df$Word1, levels = lvls) # Using this we convert both columns to factor
df$Word2 <- factor(df$Word2, levels = lvls)

df$Word1 <- as.integer(df$Word1)            # When converting this to integer, only level IDs are kept
df$Word2 <- as.integer(df$Word2)

df
#>   Word1 Word2 Count
#> 1     1     3     4
#> 2     2     1     2
#> 3     3     2     1

In igraph , tidygraph etc you also keep a second data.frame which consists of the level names (ie, the node description).

(在igraph ， tidygraph等中，您还保留了第二个data.frame ，它由级别名称（即节点描述）组成。)

We can create this from the levels saved before:

(我们可以从之前保存的级别中创建：)

df_nodes <- data.frame(names = lvls, stringsAsFactors = FALSE)
df_nodes
#>   names
#> 1     a
#> 2     c
#> 3     b

data (数据)

df <- read.csv(text = "Word1,Word2,Count
a,b,4
c,a,2
b,c,1", stringsAsFactors = FALSE)