为什么 Quanteda 频率的列/行的结果不同。共现矩阵？

Question

我正在尝试使用 Quanteda 计算一个季度内不同术语与特定术语（例如 Vietnam 或“越南”）同时出现的次数。

但是当我select来自频率共现矩阵的列或行时，计数是不同的。

谁能告诉我为什么会这样或者我做错了什么？我担心我基于这些结果的分析不正确。

##Producing the FCM

> corp <- corpus(data_SCS14q4)
> toks <- tokens(corp, remove_punct = TRUE) %>%  tokens_remove(ch_stop) %>% tokens_compound(phrase("东 盟"), concatenator = "") 
> fcm_14q4 <- fcm(toks, context = "window")

##taking the row for Vietnam or "越南":

mt <- fcm_14q4["越南",]
> head(mt)

Feature co-occurrence matrix of: 1 by 6 features.
        features
features 印 司令 中国 2050 收复 台湾
    越南  0    0    0    0    0    0

##Taking the column for Vietnam or "越南":

> mt2 <- fcm_14q4[,"越南"]
> head(mt2)

Feature co-occurrence matrix of: 6 by 1 feature.
        features
features 越南
    印      0
    司令    0
    中国   68
    2050    0
    收复    8
    台湾    4

Answer 1

这是因为默认情况下，fcm() returns只有对称共生矩阵的上三角（ordered = FALSE时对称）。要使两个索引切片相等，您需要指定 tri = FALSE.

library("quanteda")
## Package version: 3.1
## Unicode version: 13.0
## ICU version: 69.1
## Parallel computing: 12 of 12 threads used.
## See https://quanteda.io for tutorials and examples.

toks <- tokens(c("a a a b b c", "a a c e", "a c e f g"))

# default is only upper triangle
fcm(toks, context = "window", window = 2, tri = TRUE)
## Feature co-occurrence matrix of: 6 by 6 features.
##         features
## features a b c e f g
##        a 8 3 3 2 0 0
##        b 0 2 2 0 0 0
##        c 0 0 0 2 1 0
##        e 0 0 0 0 1 1
##        f 0 0 0 0 0 1
##        g 0 0 0 0 0 0

这可以使它对称，在这种情况下索引切片是相同的：

fcmat2 <- fcm(toks, context = "window", window = 2, tri = FALSE)
fcmat2
## Feature co-occurrence matrix of: 6 by 6 features.
##         features
## features a b c e f g
##        a 8 3 3 2 0 0
##        b 3 2 2 0 0 0
##        c 3 2 0 2 1 0
##        e 2 0 2 0 1 1
##        f 0 0 1 1 0 1
##        g 0 0 0 1 1 0

fcmat2[, "a"]
## Feature co-occurrence matrix of: 6 by 1 features.
##         features
## features a
##        a 8
##        b 3
##        c 3
##        e 2
##        f 0
##        g 0
t(fcmat2["a", ])
## Feature co-occurrence matrix of: 6 by 1 features.
##         features
## features a
##        a 8
##        b 3
##        c 3
##        e 2
##        f 0
##        g 0

为什么 Quanteda 频率的列/行的结果不同。共现矩阵？

Why are results different for column / row of a Quanteda freq. co-occurence matrix?

r

quanteda