Question

我有一个包含超过50,000行推文的列表。现在我已经从该列表中导出了主题标签，但是现在我已经遇到了几千行看起来像这样的主题标签

hashtag1; hashtag2; hashtag3; hashtag4

由于我想进行共同标签分析，我正在寻找一种方法将这些多个主题标签彼此连接，而不必手动将这些行转换为无向边。例如：

hashtag1; hashtag2
  hashtag1; hashtag3
  hashtag1; hashtag4
  hashtag2; hashtag3
  hashtag2; hashtag4
  hashtag3; hashtag4

那么，您是否了解如何完成此任务（例如通过R）？我是一个R-noob，甚至更少＆＃34;精通＆＃34;与其他语言，但我渴望学习。

＆＃13;

structure(list(V1 = structure(c(1L, 2L, 3L, 3L, 3L, 3L, 3L, 4L, 
5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 6L, 6L, 7L, 8L, 8L, 9L, 10L, 
10L, 10L, 10L, 10L, 10L, 10L, 11L, 11L, 11L, 11L, 12L, 12L, 13L, 
13L, 13L, 13L, 14L, 14L), .Label = c("profitkapital", "resupply", 
"robotik", "rudidutschke", "russland", "sanktionen", "sanktionieren", 
"schiller", "siegertyp", "snowden", "sockeleinkommen", "solidarity", 
"sozialismus", "sozialphilosoph"), class = "factor"), V2 = structure(c(4L, 
3L, 2L, 7L, 7L, 7L, 7L, 17L, 6L, 8L, 9L, 10L, 10L, 11L, 12L, 
13L, 18L, 18L, 1L, 15L, 15L, 14L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 
4L, 4L, 4L, 4L, 5L, 5L, 4L, 4L, 4L, 4L, 16L, 16L), .Label = c("alltag", 
"arbeit", "bbq", "bge", "blockupy", "deutschland", "digitalisierung", 
"griechenland", "grundeinkommen", "hartziv", "kenfm", "kirche", 
"kopf", "kraft", "marx", "negt", "piraten", "sanktion"), class = "factor"), 
    V3 = structure(c(1L, 3L, 2L, 4L, 4L, 4L, 4L, 4L, 5L, 4L, 
    4L, 4L, 13L, 10L, 13L, 4L, 14L, 14L, 7L, 6L, 6L, 15L, 8L, 
    8L, 8L, 8L, 8L, 8L, 8L, 1L, 1L, 1L, 1L, 12L, 12L, 11L, 11L, 
    11L, 11L, 9L, 9L), .Label = c("", "abitur", "bbqrub", "bge", 
    "brd", "brecht", "deutschen", "fsa", "grundeinkommen", "hartziv", 
    "linkezukunft", "ows", "vatikan", "widerspruch", "würde"
    ), class = "factor"), V4 = structure(c(1L, 3L, 6L, 1L, 1L, 
    1L, 1L, 1L, 8L, 1L, 2L, 1L, 9L, 5L, 9L, 10L, 4L, 4L, 7L, 
    3L, 3L, 11L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    12L, 12L, 1L, 1L, 1L, 1L, 3L, 3L), .Label = c("", "bank", 
    "bge", "eilantrag", "haarp", "job", "jobcentern", "merkel", 
    "pastor", "probleme", "super", "unibrennt"), class = "factor"), 
    V5 = structure(c(1L, 3L, 5L, 1L, 1L, 1L, 1L, 1L, 7L, 1L, 
    10L, 1L, 2L, 9L, 2L, 4L, 8L, 8L, 6L, 1L, 1L, 6L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 1L, 1L, 1L, 1L, 
    1L, 1L), .Label = c("", "bge", "bgenation", "fliegen", "geld", 
    "hartziv", "hitler", "sg", "ttip", "vorbild"), class = "factor"), 
    V6 = structure(c(1L, 5L, 2L, 1L, 1L, 1L, 1L, 1L, 6L, 1L, 
    1L, 1L, 8L, 4L, 8L, 7L, 4L, 4L, 4L, 1L, 1L, 4L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 3L, 1L, 1L, 1L, 1L, 1L, 
    1L), .Label = c("", "altersarmut", "antifa", "bge", "deeznuts", 
    "holocaust", "klatsch", "sex"), class = "factor"), V7 = structure(c(1L, 
    1L, 2L, 1L, 1L, 1L, 1L, 1L, 6L, 1L, 1L, 1L, 1L, 3L, 1L, 1L, 
    4L, 4L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 5L, 5L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("", "bge", 
    "cia", "hartz", "spanishrevolution", "wahre"), class = "factor"), 
    V8 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 1L, 
    1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 4L, 4L, 1L, 1L, 1L, 1L, 1L, 
    1L), .Label = c("", "cityoflondon", "grund", "peace"), class = "factor"), 
    V9 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 
    1L, 1L, 1L, 4L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 3L, 1L, 1L, 1L, 1L, 1L, 
    1L), .Label = c("", "bge", "occupy", "rothschild"), class = "factor"), 
    V10 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 3L, 1L, 1L, 1L, 1L, 1L, 
    1L), .Label = c("", "ard", "gezi"), class = "factor"), V11 = structure(c(1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("", "refugeeswelcome", 
    "zdf"), class = "factor"), V12 = structure(c(1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("", "nolegida", 
    "wdr"), class = "factor"), V13 = structure(c(1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("", "nopegida", 
    "swr"), class = "factor"), V14 = structure(c(1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("", "nocastor", 
    "zukunft"), class = "factor")), .Names = c("V1", "V2", "V3", 
"V4", "V5", "V6", "V7", "V8", "V9", "V10", "V11", "V12", "V13", 
"V14"), class = "data.frame", row.names = c(NA, -41L))

＆＃13;

Answer 1

您可以使用combinat尝试包combn，这会产生几个排列

library(combinat)
combn(c("hashtag1", "hashtag2", "hashtag3", "hashtag4"), 2)
     [,1]       [,2]       [,3]       [,4]       [,5]       [,6]      
[1,] "hashtag1" "hashtag1" "hashtag1" "hashtag2" "hashtag2" "hashtag3"
[2,] "hashtag2" "hashtag3" "hashtag4" "hashtag3" "hashtag4" "hashtag4"

从一行连接多个主题标签

1 个答案: