<\/p>\n
R’deki bir veri \u00e7er\u00e7evesine, her sat\u0131r\u0131n ba\u015fka bir veri \u00e7er\u00e7evesinde mevcut olup olmad\u0131\u011f\u0131n\u0131 belirten yeni bir s\u00fctun eklemek i\u00e7in a\u015fa\u011f\u0131daki s\u00f6zdizimini kullanabilirsiniz:<\/span><\/p>\n Bu \u00f6zel s\u00f6zdizimi , df1’deki her sat\u0131r\u0131n df2<\/strong> ad\u0131<\/strong> verilen ba\u015fka bir veri \u00e7er\u00e7evesinde bulunup<\/strong> bulunmad\u0131\u011f\u0131n\u0131 belirtmek i\u00e7in DO\u011eRU<\/strong> veya YANLI\u015e<\/strong> i\u00e7eren df1<\/strong> adl\u0131 veri \u00e7er\u00e7evesine, mevcut adl\u0131 bir s\u00fctun ekler.<\/span><\/p>\n A\u015fa\u011f\u0131daki \u00f6rnek, bu s\u00f6zdiziminin pratikte nas\u0131l kullan\u0131laca\u011f\u0131n\u0131 g\u00f6sterir.<\/span><\/p>\n R’de a\u015fa\u011f\u0131daki iki veri \u00e7er\u00e7evesine sahip oldu\u011fumuzu varsayal\u0131m:<\/span><\/p>\n Her sat\u0131r\u0131n ikinci veri \u00e7er\u00e7evesinde var olup olmad\u0131\u011f\u0131n\u0131 belirten ilk veri \u00e7er\u00e7evesine, mevcut<\/strong> adl\u0131 bir s\u00fctun eklemek i\u00e7in a\u015fa\u011f\u0131daki s\u00f6zdizimini kullanabiliriz:<\/span><\/p>\n Yeni s\u00fctun mevcut<\/strong> , ilk veri \u00e7er\u00e7evesindeki her sat\u0131r\u0131n ikinci veri \u00e7er\u00e7evesinde mevcut olup olmad\u0131\u011f\u0131n\u0131 g\u00f6sterir.<\/span><\/p>\n Sonu\u00e7tan \u015funlar\u0131 g\u00f6rebiliriz:<\/span><\/p>\n Ve benzeri.<\/span><\/p>\n Mevcut s\u00fctunda DO\u011eRU<\/strong> veya YANLI\u015e<\/strong> yerine 1’leri<\/strong> ve 0’lar\u0131<\/strong> g\u00f6r\u00fcnt\u00fclemek i\u00e7in is.numeric()<\/strong> i\u015flevini de kullanabilece\u011finizi unutmay\u0131n:<\/span><\/p>\n 1<\/strong> de\u011feri, ilk veri blo\u011fundaki sat\u0131r\u0131n ikincide mevcut oldu\u011funu g\u00f6sterir.<\/span><\/p>\n Tersine, 0<\/strong> de\u011feri, ilk veri \u00e7er\u00e7evesindeki sat\u0131r\u0131n ikincide mevcut olmad\u0131\u011f\u0131n\u0131 g\u00f6sterir.<\/span><\/p>\n A\u015fa\u011f\u0131daki e\u011fitimlerde R’de di\u011fer ortak g\u00f6revlerin nas\u0131l ger\u00e7ekle\u015ftirilece\u011fi a\u00e7\u0131klanmaktad\u0131r:<\/span><\/p>\n C: Birden fazla s\u00fctunun e\u015fit olup olmad\u0131\u011f\u0131 nas\u0131l kontrol edilir?<\/a> R’deki bir veri \u00e7er\u00e7evesine, her sat\u0131r\u0131n ba\u015fka bir veri \u00e7er\u00e7evesinde mevcut olup olmad\u0131\u011f\u0131n\u0131 belirten yeni bir s\u00fctun eklemek i\u00e7in a\u015fa\u011f\u0131daki s\u00f6zdizimini kullanabilirsiniz: df1$exists <- do. call (paste0, df1) %in% do. call (paste0, df2) Bu \u00f6zel s\u00f6zdizimi , df1’deki her sat\u0131r\u0131n df2 ad\u0131 verilen ba\u015fka bir veri \u00e7er\u00e7evesinde bulunup bulunmad\u0131\u011f\u0131n\u0131 belirtmek i\u00e7in DO\u011eRU veya YANLI\u015e i\u00e7eren […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[11],"tags":[],"class_list":["post-4460","post","type-post","status-publish","format-standard","hentry","category-rehber"],"yoast_head":"\n df1$exists <- do. call<\/span> (paste0, df1) %in%<\/span> do. call<\/span> (paste0, df2)\n<\/strong><\/pre>\n \u00d6rnek: R’de bir veri \u00e7er\u00e7evesindeki sat\u0131r\u0131n di\u011ferinde olup olmad\u0131\u011f\u0131n\u0131 kontrol edin<\/strong><\/h2>\n
#create first data frame\n<\/span>df1 <- data. frame<\/span> (team=c('A', 'B', 'C', 'D', 'E'),\n dots=c(12, 15, 22, 29, 24))\n\n#view first data frame\n<\/span>df1\n\n team points\n1 to 12\n2 B 15\n3 C 22\n4 D 29\n5 E 24\n\n#create second data frame\n<\/span>df2 <- data. frame<\/span> (team=c('A', 'D', 'F', 'G', 'H'),\n dots=c(12, 29, 15, 19, 10))\n\n#view second data frame\n<\/span>df2\n\n team points\n1 to 12\n2 D 29\n3 F 15\n4 G 19\n5:10 a.m.<\/span><\/span><\/strong><\/pre>\n #add new column to df1 that shows if row exists in df2\n<\/span>df1$exists <- do. call<\/span> (paste0, df1) %in%<\/span> do. call<\/span> (paste0, df2)\n\n#view updated data frame\n<\/span>df1\n\n team points exists\n1 TO 12 TRUE\n2 B 15 FALSE\n3 C 22 FALSE\n4 D 29 TRUE\n5 E 24 FALSE<\/strong><\/span><\/pre>\n\n
#add new column to df1 that shows if row exists in df2\n<\/span>df1$exists <- as. numeric<\/span> (do. call<\/span> (paste0, df1) %in%<\/span> do. call<\/span> (paste0, df2))\n\n#view updated data frame\n<\/span>df1\n\n team points exists\n1 to 12 1\n2 B 15 0\n3 C 22 0\n4 D 29 1\n5 E 24 0\n<\/strong><\/span><\/pre>\n Ek kaynaklar<\/strong><\/span><\/h2>\n
C: Bir veri \u00e7er\u00e7evesindeki tek sat\u0131rlar nas\u0131l se\u00e7ilir?<\/a>
A: Bir veri \u00e7er\u00e7evesindeki sat\u0131rlar nas\u0131l \u00e7o\u011falt\u0131l\u0131r<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"