[EDIT2]

Question

我试图以字符串系数执行字词计数。

我有一个名为Names的因子。以下示例：

[1] GP - Hyperion Planning Upgrade
[2] Application Support Renewal
[3] Oracle EBS upgrade 11.5 to 12.1
[4] Bluemix Services
[5] Cognos 11 Upgrade

我还有一个名为terms的词典术语列表：

[1] "IBM"     "Oracle"     "SQL Server"     "Cognos"     "Azure"

我需要R从terms列表中创建一个数据框，并从Names因子创建每个团队的总数。例如：

         term       count
1        IBM         3
2        Oracle      6
3        SQL Server  0
4        Cognos      2
5        Azure       9

值得注意的是：该术语可以在一个名称中多次出现。它算作一次。

我想问一下，是否有人可以从中得到任何关于此的例子。感谢。

Answer 1

您可以尝试此操作（稍微更改向量Names并假设您需要不区分大小写的匹配项）：

# input
Names <- as.character(Names)
Names
#[1] IBM GP - Hyperion IBM Planning Upgrade IBM"
#[2] Application Support Renewal"               
#[3] Oracle EBS upgrade 11.5 to 12.1"           
#[4] Bluemix Services IBM"                      
#[5] Cognos 11 Upgrade"  

terms <- c("IBM",     "Oracle",     "SQL Server",     "Cognos",     "Azure")
vgrepl <- Vectorize(grepl, 'pattern', SIMPLIFY = TRUE)
df <- +(vgrepl(tolower(terms), tolower(Names))) # case insensitive

df
#     ibm oracle sql server cognos azure
#[1,]   1      0          0      0     0
#[2,]   0      0          0      0     0
#[3,]   0      1          0      0     0
#[4,]   1      0          0      0     0
#[5,]   0      0          0      1     0

colSums(df)
#    ibm     oracle sql server     cognos      azure 
#     2          1          0          1          0 

data.frame(count=colSums(df))
#           count
#ibm            2
#oracle         1
#sql server     0
#cognos         1
#azure          0

[EDIT2]

df <- data.frame(count=colSums(df))
df <- cbind.data.frame(terms=rownames(df), df)
df
#                terms count
#ibm               ibm     2
#oracle         oracle     1
#sql server sql server     0
#cognos         cognos     1
#azure           azure     0

Answer 2

这是一个避免正则表达式支持match的示例：

names <- c(
"GP - Hyperion Planning Upgrade",
"Application Support Renewal",
"Oracle EBS upgrade 11.5 to 12.1",
"Bluemix Services",
"Cognos 11 Upgrade")

terms <- tolower(c("IBM", "Oracle", "SQL Server", "Cognos", "Azure"))

## Split your names on whitespace and match each token to the terms
counts <- lapply(strsplit(tolower(names), "\\s+"), match, terms, 0)

## index the terms using the match indices and table it
table(terms[unlist(counts)])

cognos oracle 
     1      1

如何根据r中的字典术语列表计算数据框中的单词

2 个答案:

[EDIT2]