{"id":12095,"date":"2022-09-29T14:44:55","date_gmt":"2022-09-29T09:14:55","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-finding-the-index-based-on-two-data-frames-of-strings\/"},"modified":"2022-09-29T14:44:55","modified_gmt":"2022-09-29T09:14:55","slug":"solved-finding-the-index-based-on-two-data-frames-of-strings","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-finding-the-index-based-on-two-data-frames-of-strings\/","title":{"rendered":"[Solved] Finding the index based on two data frames of strings"},"content":{"rendered":"

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A possible solution with base R by using a combination of colSums<\/code>, which<\/code>, toString<\/code> and apply<\/code>:<\/p>\n

strs$colids <- apply(strs, 1, function(x) toString(which(colSums(lut == x, na.rm=TRUE) > 0)))\n<\/code><\/pre>\n
which gives:<\/p>\n
> strs\n  strings  colids\n1  O75663    1, 3\n2  O95400    1, 3\n3  O95433    1, 3\n4  O95456 2, 3, 4\n5  O95670 2, 3, 4\n6  O95801       4\n7  P00352       4\n8  P00492        \n<\/code><\/pre>\n
To see what each part does, start by looking at the output of lut == 'O75663'<\/code> which will give you a TRUE\/FALSE<\/code> table. By wrapping this in colSums<\/code> you sum the TRUE\/FALSE<\/code>. A 0<\/code> means that there is no match in that column for that string, a number above zero means that there is one or more matches. With which<\/code> you get the column indexes and by wrapping that in toString<\/code> you get a character values with indexes of the matching columns.<\/p>\n
This approach could also be implemented with either data.table<\/code> or dplyr<\/code>:<\/p>\n
library(data.table)\nsetDT(strs)[, colids := toString(which(colSums(lut == strings, na.rm=TRUE) > 0)), by = 1:nrow(strs)][]\n\nlibrary(dplyr)\nstrs %>% rowwise() %>% mutate(colids = toString(which(colSums(lut == strings, na.rm=TRUE) > 0)))\n<\/code><\/pre>\n
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In response to your comment: An example for multiple columns in strs<\/code> with data.table<\/code>:<\/p>\n
# create an extra strings column\nset.seed(1)\nstrs$strings2 <- sample(strs$strings)\n\n# create two 'colids' columns\nlibrary(data.table)\nsetDT(strs)[, c('colids1','colids2') := lapply(.SD, function(x) toString(which(colSums(lut == x, na.rm=TRUE) > 0))), by = 1:nrow(strs)][]\n<\/code><\/pre>\n
which gives:<\/p>\n
   strings strings2 colids1 colids2\n1:  O75663   O95433    1, 3    1, 3\n2:  O95400   P00492    1, 3        \n3:  O95433   O95456    1, 3 2, 3, 4\n4:  O95456   O95670 2, 3, 4 2, 3, 4\n5:  O95670   O75663 2, 3, 4    1, 3\n6:  O95801   P00352       4       4\n7:  P00352   O95400       4    1, 3\n8:  P00492   O95801               4\n<\/code><\/pre>\n
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Used data<\/em>:<\/p>\n
lut <- structure(list(V1 = c(\"O75663\", \"O95400\", \"O95433\", NA, NA), \n                      V2 = c(\"O95456\", \"O95670\", NA, NA, NA), \n                      V3 = c(\"O75663\", \"O95400\", \"O95433\", \"O95456\", \"O95670\"), \n                      V4 = c(\"O95456\", \"O95670\", \"O95801\", \"P00352\", NA)), \n                 .Names = c(\"V1\", \"V2\", \"V3\", \"V4\"), class = \"data.frame\", row.names = c(NA, -5L))\n\nstrs <- structure(list(strings = c(\"O75663\", \"O95400\", \"O95433\", \"O95456\", \"O95670\", \"O95801\", \"P00352\", \"P00492\")), \n                  .Names = \"strings\", class = \"data.frame\", row.names = c(NA, -8L))\n<\/code><\/pre>\n
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With regard to the extended example you added to your question: The reason you are getting that error is because you are trying to compare factor-variables with character-variables. See the difference in output between sapply(strs,class)<\/code> and sapply(lut,class)<\/code>:<\/p>\n
> sapply(strs,class)\nstrings1 strings2 strings3 strings4 strings5 \n\"factor\" \"factor\" \"factor\" \"factor\" \"factor\" \n> sapply(lut,class)\n         V1          V2          V3          V4          V5          V6          V7          V8 \n\"character\" \"character\" \"character\" \"character\" \"character\" \"character\" \"character\" \"character\" \n<\/code><\/pre>\n
It is therefore necessary to convert the factor<\/code>‘s to character<\/code>‘s first and then do the comparison. The following code:<\/p>\n
library(data.table)\nsetDT(strs)[, lapply(.SD, as.character)\n            ][, paste0('colids.',seq_along(strs)) := lapply(.SD, function(x) toString(which(colSums(lut == x, na.rm=TRUE) > 0))), \n              by = 1:nrow(strs)][]\n<\/code><\/pre>\n
now gives the correct output:<\/p>\n
   strings1 strings2 strings3 strings4 strings5         colids.1         colids.2         colids.3         colids.4         colids.5\n1:   O75663   O95456   O95456   O95400   P00492       1, 3, 5, 7 2, 3, 4, 6, 7, 8 2, 3, 4, 6, 7, 8       1, 3, 5, 7                 \n2:   O95400   O75663   O95801   P00492   O95400       1, 3, 5, 7       1, 3, 5, 7             4, 8                        1, 3, 5, 7\n3:   O95433   P00492   P00352   O95456   P00352       1, 3, 5, 7                              4, 8 2, 3, 4, 6, 7, 8             4, 8\n4:   O95456   P00352   P00492   O95801   O75663 2, 3, 4, 6, 7, 8             4, 8                              4, 8       1, 3, 5, 7\n5:   O95670   O95433   O75663            O95433 2, 3, 4, 6, 7, 8       1, 3, 5, 7       1, 3, 5, 7                        1, 3, 5, 7\n6:   O95801            O95400            O95801             4, 8                        1, 3, 5, 7                              4, 8\n7:                     O95670            O95670                                   2, 3, 4, 6, 7, 8                  2, 3, 4, 6, 7, 8\n8:                                       O95456                                                                     2, 3, 4, 6, 7, 8\n<\/code><\/pre>\n
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Used data extended example<\/em>:<\/p>\n
strs <- structure(list(strings1 = structure(c(2L, 3L, 4L, 5L, 6L, 7L, 1L, 1L), .Label = c(\"\", \"O75663\", \"O95400\", \"O95433\", \"O95456\", \"O95670\", \"O95801\"), class = \"factor\"), \n                       strings2 = structure(c(4L, 2L, 6L, 5L, 3L, 1L, 1L, 1L), .Label = c(\"\", \"O75663\", \"O95433\", \"O95456\", \"P00352\", \"P00492\"), class = \"factor\"), \n                       strings3 = structure(c(4L, 6L, 7L, 8L, 2L, 3L, 5L, 1L), .Label = c(\"\", \"O75663\", \"O95400\", \"O95456\", \"O95670\", \"O95801\", \"P00352\", \"P00492\"), class = \"factor\"), \n                       strings4 = structure(c(2L, 5L, 3L, 4L, 1L, 1L, 1L, 1L), .Label = c(\"\", \"O95400\", \"O95456\", \"O95801\", \"P00492\"), class = \"factor\"), \n                       strings5 = structure(c(8L, 2L, 7L, 1L, 3L, 6L, 5L, 4L), .Label = c(\"O75663\", \"O95400\", \"O95433\", \"O95456\", \"O95670\", \"O95801\", \"P00352\", \"P00492\"), class = \"factor\")), \n                  .Names = c(\"strings1\", \"strings2\", \"strings3\", \"strings4\", \"strings5\"), class = \"data.frame\", row.names = c(NA, -8L))\n\nlut <- structure(list(V1 = c(\"O75663\", \"O95400\", \"O95433\", NA, NA), \n                      V2 = c(\"O95456\", \"O95670\", NA, NA, NA), \n                      V3 = c(\"O75663\", \"O95400\", \"O95433\", \"O95456\", \"O95670\"), \n                      V4 = c(\"O95456\", \"O95670\", \"O95801\", \"P00352\", NA), \n                      V5 = c(\"O75663\", \"O95400\", \"O95433\", NA, NA), \n                      V6 = c(\"O95456\", \"O95670\", NA, NA, NA), \n                      V7 = c(\"O75663\", \"O95400\", \"O95433\", \"O95456\", \"O95670\"), \n                      V8 = c(\"O95456\", \"O95670\", \"O95801\", \"P00352\", NA)), \n                 .Names = c(\"V1\", \"V2\", \"V3\", \"V4\", \"V5\", \"V6\", \"V7\", \"V8\"), row.names = c(NA, -5L), class = \"data.frame\")\n<\/code><\/pre>\n<\/p><\/div>\n
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            14<\/span> <\/p><\/div>\n<\/div>\n
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solved Finding the index based on two data frames of strings <\/p>\n","protected":false},"excerpt":{"rendered":"
[ad_1] A possible solution with base R by using a combination of colSums, which, toString and apply: strs$colids <- apply(strs, 1, function(x) toString(which(colSums(lut == x, na.rm=TRUE) > 0))) which gives: > strs strings colids 1 O75663 1, 3 2 O95400 1, 3 3 O95433 1, 3 4 O95456 2, 3, 4 5 O95670 2, 3, … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[320],"tags":[321],"class_list":["post-12095","post","type-post","status-publish","format-standard","hentry","category-solved","tag-r"],"yoast_head":"\n[Solved] Finding the index based on two data frames of strings - JassWeb<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/jassweb.com\/solved\/solved-finding-the-index-based-on-two-data-frames-of-strings\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] Finding the index based on two data frames of strings - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] A possible solution with base R by using a combination of colSums, which, toString and apply: strs$colids <- apply(strs, 1, function(x) toString(which(colSums(lut == x, na.rm=TRUE) > 0))) which gives: > strs strings colids 1 O75663 1, 3 2 O95400 1, 3 3 O95433 1, 3 4 O95456 2, 3, 4 5 O95670 2, 3, ... 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