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R:使用plyr在两个数据源的匹配子集之间进行模糊字符串匹配

转载 作者:行者123 更新时间:2023-12-03 03:02:55 24 4
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假设我有一个县列表,其中存在不同数量的拼写错误或其他问题,这些问题将它们与 2010 FIPS dataset 区分开来。 (下面用于创建 fips 数据框的代码),但拼写错误的县所在的州已正确输入。以下是我的完整数据集中 21 个随机观察结果的样本:

tomatch <- structure(list(county = c("Beauregard", "De Soto", "Dekalb", "Webster",
"Saint Joseph", "West Feliciana", "Ketchikan Gateway", "Evangeline",
"Richmond City", "Saint Mary", "Saint Louis City", "Mclean",
"Union", "Bienville", "Covington City", "Martinsville City",
"Claiborne", "King And Queen", "Mclean", "Mcminn", "Prince Georges"
), state = c("LA", "LA", "GA", "LA", "IN", "LA", "AK", "LA", "VA",
"LA", "MO", "KY", "LA", "LA", "VA", "VA", "LA", "VA", "ND", "TN",
"MD")), .Names = c("county", "state"), class = c("tbl_df", "data.frame"
), row.names = c(NA, -21L))

county state
1 Beauregard LA
2 De Soto LA
3 Dekalb GA
4 Webster LA
5 Saint Joseph IN
6 West Feliciana LA
7 Ketchikan Gateway AK
8 Evangeline LA
9 Richmond City VA
10 Saint Mary LA
11 Saint Louis City MO
12 Mclean KY
13 Union LA
14 Bienville LA
15 Covington City VA
16 Martinsville City VA
17 Claiborne LA
18 King And Queen VA
19 Mclean ND
20 Mcminn TN
21 Prince Georges MD

我使用 adist 创建了一个模糊字符串匹配算法,该算法将大约 80% 的县与 fips 中的县名称相匹配。但是,有时它会匹配两个拼写相似但来自不同州的县(例如,“Webster, LA”匹配“Webster, GA”而不是“Webster Parrish, LA”)。

distance <- adist(tomatch$county, 
fips$countyname,
partial = TRUE)


min.name <- apply(distance, 1, min)

matchedcounties <- NULL

for(i in 1:nrow(distance)) {

s2.i <- match(min.name[i], distance[i, ])
s1.i <- i

matchedcounties <- rbind(data.frame(s2.i = s2.i,
s1.i = s1.i,
s1name = tomatch[s1.i, ]$county,
s2name = fips[s2.i, ]$countyname,
adist = min.name[i]),
matchedcounties)

}

因此,我想将县的模糊字符串匹配限制为具有匹配州的拼写正确的版本。

我当前的算法创建一个大矩阵,计算两个源之间的标准 Levenshtein 距离,然后选择距离最小的值。

为了解决我的问题,我猜我需要创建一个可以通过 ddply 应用于每个“状态”组的函数,但我很困惑我应该如何做指示 ddply 函数中的组值应与另一个数据帧匹配。 dplyr 解决方案或使用任何其他包的解决方案也将受到赞赏。

创建 FIPS 数据集的代码:

download.file('http://www2.census.gov/geo/docs/reference/codes/files/national_county.txt',
'./nationalfips.txt')

fips <- read.csv('./nationalfips.txt',
stringsAsFactors = FALSE, colClasses = 'character', header = FALSE)
names(fips) <- c('state', 'statefips', 'countyfips', 'countyname', 'classfips')

# remove 'County' from countyname
fips$countyname <- sub('County', '', fips$countyname, fixed = TRUE)
fips$countyname <- stringr::str_trim(fips$countyname)

最佳答案

这是 dplyr 的一种方法。我首先按州加入 tomatch data.frame 与 FIPS 名称(仅允许州内匹配):

require(dplyr)
df <- tomatch %>%
left_join(fips, by="state")

接下来,我注意到很多县没有“Saint”,而是“St.”在 FIPS 数据集中。首先清理它应该可以改善获得的结果。

df <- df %>%
mutate(county_clean = gsub("Saint", "St.", county))

然后,按县对这个 data.frame 进行分组,并使用 adist 计算距离:

df <- df %>%
group_by(county_clean) %>% # Calculate the distance per county
mutate(dist = diag(adist(county_clean, countyname, partial=TRUE))) %>%
arrange(county, dist) # Used this for visual inspection.

请注意,我从结果矩阵中取出了对角线,因为 adist 返回一个 n x m 矩阵,其中 n 代表 x 向量,m 代表 y 向量(它计算所有组合)。或者,您可以添加 agrep 结果:

df <- df %>%
rowwise() %>% # 'group_by' a single row.
mutate(agrep_result = agrepl(county_clean, countyname, max.distance = 0.3)) %>%
ungroup() # Always a good idea to remove 'groups' after you're done.

然后像之前一样进行过滤,取最小距离:

df <- df %>%
group_by(county_clean) %>% # Causes it to calculate the 'min' per group
filter(dist == min(dist)) %>%
ungroup()

请注意,这可能会导致 tomatch 中的每一输入行返回多行。
或者,一次运行即可完成所有操作(一旦我确信代码正在执行其应该执行的操作,我通常会将其更改为这种格式):

df <- tomatch %>% 
# Join on all names in the relevant state and clean 'St.'
left_join(fips, by="state") %>%
mutate(county_clean = gsub("Saint", "St.", county)) %>%

# Calculate the distances, per original county name.
group_by(county_clean) %>%
mutate(dist = diag(adist(county_clean, countyname, partial=TRUE))) %>%

# Append the agrepl result
rowwise() %>%
mutate(string_agrep = agrepl(county_clean, countyname, max.distance = 0.3)) %>%
ungroup() %>%

# Only retain minimum distances
group_by(county_clean) %>%
filter(dist == min(dist))

两种情况的结果:

              county      county_clean state                countyname dist string_agrep
1 Beauregard Beauregard LA Beauregard Parish 0 TRUE
2 De Soto De Soto LA De Soto Parish 0 TRUE
3 Dekalb Dekalb GA DeKalb 1 TRUE
4 Webster Webster LA Webster Parish 0 TRUE
5 Saint Joseph St. Joseph IN St. Joseph 0 TRUE
6 West Feliciana West Feliciana LA West Feliciana Parish 0 TRUE
7 Ketchikan Gateway Ketchikan Gateway AK Ketchikan Gateway Borough 0 TRUE
8 Evangeline Evangeline LA Evangeline Parish 0 TRUE
9 Richmond City Richmond City VA Richmond city 1 TRUE
10 Saint Mary St. Mary LA St. Mary Parish 0 TRUE
11 Saint Louis City St. Louis City MO St. Louis city 1 TRUE
12 Mclean Mclean KY McLean 1 TRUE
13 Union Union LA Union Parish 0 TRUE
14 Bienville Bienville LA Bienville Parish 0 TRUE
15 Covington City Covington City VA Covington city 1 TRUE
16 Martinsville City Martinsville City VA Martinsville city 1 TRUE
17 Claiborne Claiborne LA Claiborne Parish 0 TRUE
18 King And Queen King And Queen VA King and Queen 1 TRUE
19 Mclean Mclean ND McLean 1 TRUE
20 Mcminn Mcminn TN McMinn 1 TRUE
21 Prince Georges Prince Georges MD Prince George's 1 TRU

关于R:使用plyr在两个数据源的匹配子集之间进行模糊字符串匹配,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/30898630/

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