我正在通过quanteda包在字典中读取R。这个软件包预装了一些很棒的词典，其中之一是我感兴趣的道德基础词典。这本词典有几个类别(农场，公平，内组等(，分为美德和副子类别。
我想计算 R 中每个基础的每个子类别中的单词数。我该怎么做呢？
对于一个可重现的例子，我可以通过运行library(quanteda.dictionaries)来访问道德基础词典(标记为data_dictionary_MFD
谢谢！
目前还不完全清楚您要查找的内容，但这可能归结为术语。quanteda字典使用术语"键"表示规范类别(在 R 中为列表元素的名称(，使用术语"值"表示用于匹配单词以计算每个键的出现次数的模式。
MFD有两套"钥匙"：道德"基础"，如关怀，公平等，以及每个基础类别的"恶习"和"美德"所代表的"价"。 正如我们在quanteda.dictionaries::data_dictionary_MFD中记录的那样 - 至少在quanteda.dictionary的v0.22中 - 字典被扁平化到一个级别。
我们可以看到这一点，并计算每个字典"键"中的值，该字典"键"在这里结合了基础和价，如下所示：
library("quanteda")
## Package version: 1.5.2
data(data_dictionary_MFD, package = "quanteda.dictionaries")
# number of "words" in each MFD dictionary key
lengths(data_dictionary_MFD)
## care.virtue care.vice fairness.virtue fairness.vice
## 182 288 115 236
## loyalty.virtue loyalty.vice authority.virtue authority.vice
## 142 49 301 130
## sanctity.virtue sanctity.vice
## 272 388
# first 5 values in each dictionary key
lapply(data_dictionary_MFD, head, 5)
## $care.virtue
## [1] "alleviate" "alleviated" "alleviates" "alleviating" "alleviation"
##
## $care.vice
## [1] "abused" "abuser" "abusers" "abuses" "abusing"
##
## $fairness.virtue
## [1] "avenge" "avenged" "avenger" "avengers" "avenges"
##
## $fairness.vice
## [1] "am partial" "bamboozle" "bamboozled" "bamboozles" "bamboozling"
##
## $loyalty.virtue
## [1] "all for one" "allegiance" "allegiances" "allegiant" "allied"
##
## $loyalty.vice
## [1] "against us" "apostate" "apostates" "backstab" "backstabbed"
##
## $authority.virtue
## [1] "acquiesce" "acquiesced" "acquiescent" "acquiesces" "acquiescing"
##
## $authority.vice
## [1] "anarchist" "anarchistic" "anarchists" "anarchy" "apostate"
##
## $sanctity.virtue
## [1] "abstinance" "abstinence" "allah" "almighty" "angel"
##
## $sanctity.vice
## [1] "abhor" "abhored" "abhors" "addict" "addicted"
要应用它来计算与"键"匹配的单词(基础和价的组合(，我们可以创建一个 dfm，然后使用dfm_lookup()：
# number of words in a text matching the MFD dictionary
dfm(data_corpus_inaugural) %>%
dfm_lookup(dictionary = data_dictionary_MFD) %>%
tail()
## Document-feature matrix of: 6 documents, 10 features (10.0% sparse).
## 6 x 10 sparse Matrix of class "dfm"
## features
## docs care.virtue care.vice fairness.virtue fairness.vice
## 1997-Clinton 8 4 6 2
## 2001-Bush 21 8 11 1
## 2005-Bush 14 12 16 4
## 2009-Obama 18 6 8 1
## 2013-Obama 14 6 15 2
## 2017-Trump 16 7 2 4
## features
## docs loyalty.virtue loyalty.vice authority.virtue authority.vice
## 1997-Clinton 37 0 3 0
## 2001-Bush 36 1 18 2
## 2005-Bush 38 3 33 4
## 2009-Obama 33 1 18 2
## 2013-Obama 39 2 12 0
## 2017-Trump 44 0 20 1
## features
## docs sanctity.virtue sanctity.vice
## 1997-Clinton 14 8
## 2001-Bush 21 1
## 2005-Bush 16 0
## 2009-Obama 18 3
## 2013-Obama 14 0
## 2017-Trump 13 3
但是，有一种更好的方法可以利用MFD的嵌套结构，但是我们需要先修改字典对象以使其嵌套。 如前所述，MFD 已经"扁平化"。 我们想把它放平，以便基础形成第一级键，化合价形成第二级键。 然后，使用levels中的参数tokens_lookup()和dfm_lookup()，我们将能够选择在文本中计算匹配项的级别。
首先，重新创建字典以使其嵌套。
# remake the dictionary into nested catetgory of foundation and valence
data_dictionary_MFDnested <-
dictionary(list(
care = list(
virtue = data_dictionary_MFD[["care.virtue"]],
vice = data_dictionary_MFD[["care.vice"]]
),
fairness = list(
virtue = data_dictionary_MFD[["fairness.virtue"]],
vice = data_dictionary_MFD[["fairness.vice"]]
),
loyalty = list(
virtue = data_dictionary_MFD[["loyalty.virtue"]],
vice = data_dictionary_MFD[["loyalty.vice"]]
),
authority = list(
virtue = data_dictionary_MFD[["authority.virtue"]],
vice = data_dictionary_MFD[["authority.vice"]]
),
sanctity = list(
virtue = data_dictionary_MFD[["sanctity.virtue"]],
vice = data_dictionary_MFD[["sanctity.vice"]]
)
))
检查这一点，我们可以看到字典上的详细信息：
lengths(data_dictionary_MFDnested)
## care fairness loyalty authority sanctity
## 2 2 2 2 2
lapply(data_dictionary_MFDnested, lengths)
## $care
## virtue vice
## 182 288
##
## $fairness
## virtue vice
## 115 236
##
## $loyalty
## virtue vice
## 142 49
##
## $authority
## virtue vice
## 301 130
##
## $sanctity
## virtue vice
## 272 388
现在我们可以将其应用于我们的文本：
# now apply it to texts
dfm(data_corpus_inaugural) %>%
dfm_lookup(dictionary = data_dictionary_MFDnested, levels = 1) %>%
tail()
## Document-feature matrix of: 6 documents, 5 features (0.0% sparse).
## 6 x 5 sparse Matrix of class "dfm"
## features
## docs care fairness loyalty authority sanctity
## 1997-Clinton 12 8 37 3 22
## 2001-Bush 29 12 37 20 22
## 2005-Bush 26 20 41 37 16
## 2009-Obama 24 9 34 20 21
## 2013-Obama 20 17 41 12 14
## 2017-Trump 23 6 44 21 16
dfm(data_corpus_inaugural) %>%
dfm_lookup(dictionary = data_dictionary_MFDnested, levels = 2) %>%
tail()
## Document-feature matrix of: 6 documents, 2 features (0.0% sparse).
## 6 x 2 sparse Matrix of class "dfm"
## features
## docs virtue vice
## 1997-Clinton 68 14
## 2001-Bush 107 13
## 2005-Bush 117 23
## 2009-Obama 95 13
## 2013-Obama 94 10
## 2017-Trump 95 15
指定两个级别(或默认值levels = 1:5(与我们最初使用扁平字典的内容相匹配：
dfm(data_corpus_inaugural) %>%
dfm_lookup(dictionary = data_dictionary_MFDnested, levels = 1:2) %>%
tail()
## Document-feature matrix of: 6 documents, 10 features (10.0% sparse).
## 6 x 10 sparse Matrix of class "dfm"
## features
## docs care.virtue care.vice fairness.virtue fairness.vice
## 1997-Clinton 8 4 6 2
## 2001-Bush 21 8 11 1
## 2005-Bush 14 12 16 4
## 2009-Obama 18 6 8 1
## 2013-Obama 14 6 15 2
## 2017-Trump 16 7 2 4
## features
## docs loyalty.virtue loyalty.vice authority.virtue authority.vice
## 1997-Clinton 37 0 3 0
## 2001-Bush 36 1 18 2
## 2005-Bush 38 3 33 4
## 2009-Obama 33 1 18 2
## 2013-Obama 39 2 12 0
## 2017-Trump 44 0 20 1
## features
## docs sanctity.virtue sanctity.vice
## 1997-Clinton 14 8
## 2001-Bush 21 1
## 2005-Bush 16 0
## 2009-Obama 18 3
## 2013-Obama 14 0
## 2017-Trump 13 3
不确定您的MFD语料库是什么样子的;如果它是托管在 osf.io/whjt2 上的语料库，那么前六行将如下所示(mfd作为数据集的名称，Wordtoken和MFDcategory作为我的列标题(：
head(mfd)
Wordtoken MFDcategory
1 compassion 1
2 empathy 1
3 kindness 1
4 caring 1
5 generosity 1
6 benevolence 1
如果您的目标只是找出在MFDcategory的十个级别中每个级别下列出了多少个单词，那么您所要做的就是对该列使用table：
table(mfd$MFDcategory)
1 2 3 4 5 6 7 8 9 10
182 288 115 236 143 49 301 130 272 388
也就是说，类别1 有 182 个单词标记，即 care.virtue，而类别 2 有 288 个单词标记，即 care.vice，依此类推。这有帮助吗？