<\/p>\n
\u00c7o\u011fu zaman, R’deki bir veri \u00e7er\u00e7evesinin s\u00fctunlar\u0131ndaki eksik de\u011ferleri, s\u00f6z konusu s\u00fctunun ortalamas\u0131 veya medyan\u0131 ile de\u011fi\u015ftirmek isteyebilirsiniz.<\/span><\/p>\n Tek bir s\u00fctundaki eksik de\u011ferleri de\u011fi\u015ftirmek i\u00e7in a\u015fa\u011f\u0131daki s\u00f6zdizimini kullanabilirsiniz:<\/span><\/p>\n Birden fazla s\u00fctundaki eksik de\u011ferleri de\u011fi\u015ftirmek i\u00e7in a\u015fa\u011f\u0131daki s\u00f6zdizimini kullanabilirsiniz:<\/span><\/p>\n Bu e\u011fitimde bu i\u015flevlerin pratikte nas\u0131l kullan\u0131laca\u011f\u0131 tam olarak a\u00e7\u0131klanmaktad\u0131r.<\/span><\/p>\n A\u015fa\u011f\u0131daki kod, bir veri \u00e7er\u00e7evesinin ilk s\u00fctunundaki eksik de\u011ferlerin, ilk s\u00fctunun ortalama de\u011feriyle nas\u0131l de\u011fi\u015ftirilece\u011fini g\u00f6sterir:<\/span><\/p>\n \u0130lk s\u00fctundaki ortalama de\u011fer 3,333<\/strong> oldu\u011fundan, ilk s\u00fctundaki eksik de\u011ferler 3,333<\/strong> ile de\u011fi\u015ftirildi.<\/span><\/p>\n A\u015fa\u011f\u0131daki kod, her s\u00fctundaki eksik de\u011ferlerin kendi s\u00fctununun ortalamas\u0131yla nas\u0131l de\u011fi\u015ftirilece\u011fini g\u00f6sterir:<\/span><\/p>\n A\u015fa\u011f\u0131daki kod, bir veri \u00e7er\u00e7evesinin ilk s\u00fctunundaki eksik de\u011ferlerin ilk s\u00fctunun medyan de\u011feriyle nas\u0131l de\u011fi\u015ftirilece\u011fini g\u00f6sterir:<\/span><\/p>\n \u0130lk s\u00fctundaki medyan de\u011feri 4<\/strong> oldu\u011fundan, ilk s\u00fctundaki eksik de\u011ferler 4<\/strong> ile de\u011fi\u015ftirildi.<\/span><\/p>\n A\u015fa\u011f\u0131daki kod, her s\u00fctundaki eksik de\u011ferlerin kendi s\u00fctununun medyan\u0131 ile nas\u0131l de\u011fi\u015ftirilece\u011fini g\u00f6sterir:<\/span><\/p>\n R’de s\u00fctun adlar\u0131 \u00fczerinde yineleme nas\u0131l yap\u0131l\u0131r<\/a> \u00c7o\u011fu zaman, R’deki bir veri \u00e7er\u00e7evesinin s\u00fctunlar\u0131ndaki eksik de\u011ferleri, s\u00f6z konusu s\u00fctunun ortalamas\u0131 veya medyan\u0131 ile de\u011fi\u015ftirmek isteyebilirsiniz. Tek bir s\u00fctundaki eksik de\u011ferleri de\u011fi\u015ftirmek i\u00e7in a\u015fa\u011f\u0131daki s\u00f6zdizimini kullanabilirsiniz: df$col[ is.na (df$col)] <- mean(df$col, na.rm = TRUE ) Birden fazla s\u00fctundaki eksik de\u011ferleri de\u011fi\u015ftirmek i\u00e7in a\u015fa\u011f\u0131daki s\u00f6zdizimini kullanabilirsiniz: for(i in 1: ncol (df)) { df[ , […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[11],"tags":[],"class_list":["post-1094","post","type-post","status-publish","format-standard","hentry","category-rehber"],"yoast_head":"\n df$col[ is.na<\/span> (df$col)] <- mean(df$col, na.rm<\/span> = TRUE<\/span> )\n<\/strong><\/pre>\n for(i in 1: ncol<\/span> (df)) {\n df[ , i][ is.na<\/span> (df[ , i])] <- mean(df[ , i], na.rm<\/span> = TRUE<\/span> )\n}\n<\/strong><\/pre>\n \u00d6rnek 1: Eksik de\u011ferleri s\u00fctun ortalamalar\u0131yla de\u011fi\u015ftirin<\/strong><\/span><\/h3>\n
#create data frame<\/span>\ndf <- data.frame(var1=c(1, NA, NA, 4, 5),\n var2=c(7, 7, 8, 3, 2),\n var3=c(3, 3, 6, 6, 8),\n var4=c(1, 1, 2, 8, 9))\n\n#replace missing values in first column with mean of first column<\/span>\ndf$var1[ is.na<\/span> (df$var1)] <- mean(df$var1, na.rm<\/span> = TRUE<\/span> )\n\n#view data frame with missing values replaced<\/span>\ndf\n\n var1 var2 var3 var4\n1 1.000000 7 3 1\n2 3.333333 7 3 1\n3 3.333333 8 6 2\n4 4.000000 3 6 8\n5 5.000000 2 8 9\n<\/strong><\/pre>\n #create data frame<\/span>\ndf <- data.frame(var1=c(1, NA, NA, 4, 5),\n var2=c(7, 7, 8, NA, 2),\n var3=c(NA, 3, 6, NA, 8),\n var4=c(1, 1, 2, 8, 9))\n\n#replace missing values in each column with column means<\/span>\nfor(i in 1: ncol<\/span> (df)) {\n df[ , i][ is.na<\/span> (df[ , i])] <- mean(df[ , i], na.rm<\/span> = TRUE<\/span> )\n}\n\n#view data frame with missing values replaced<\/span>\ndf\n\n var1 var2 var3 var4\n1 1.000000 7 5.666667 1\n2 3.333333 7 3.000000 1\n3 3.333333 8 6.000000 2\n4 4.000000 6 5.666667 8\n5 5.000000 2 8.000000 9<\/strong><\/pre>\n \u00d6rnek 2: Eksik de\u011ferleri s\u00fctun medyanlar\u0131yla de\u011fi\u015ftirin<\/strong><\/span><\/h3>\n
#create data frame<\/span>\ndf <- data.frame(var1=c(1, NA, NA, 4, 5),\n var2=c(7, 7, 8, NA, 2),\n var3=c(NA, 3, 6, NA, 8),\n var4=c(1, 1, 2, 8, 9))\n\n#replace missing values in first column with median of first column<\/span>\ndf$var1[ is.na<\/span> (df$var1)] <- median(df$var1, na.rm<\/span> = TRUE<\/span> )\n\n#view data frame with missing values replaced<\/span>\ndf\n\n var1 var2 var3 var4\n1 1 7 NA 1\n2 4 7 3 1\n3 4 8 6 2\n4 4 NA NA 8\n5 5 2 8 9\n<\/strong><\/pre>\n #create data frame<\/span>\ndf <- data.frame(var1=c(1, NA, NA, 4, 5),\n var2=c(7, 7, 8, NA, 2),\n var3=c(NA, 3, 6, NA, 8),\n var4=c(1, 1, 2, 8, 9))\n\n#replace missing values in each column with column medians<\/span>\nfor(i in 1: ncol<\/span> (df)) {\n df[ , i][ is.na<\/span> (df[ , i])] <- median(df[ , i], na.rm<\/span> = TRUE<\/span> )\n}\n\n#view data frame with missing values replaced<\/span>\ndf\n\n var1 var2 var3 var4\n1 1 7 6 1\n2 4 7 3 1\n3 4 8 6 2\n4 4 7 6 8\n5 5 2 8 9<\/strong><\/pre>\n Ek kaynaklar<\/strong><\/span><\/h3>\n
R’de birden fazla s\u00fctunun ortalamas\u0131 nas\u0131l hesaplan\u0131r<\/a>
R’de belirli s\u00fctunlar nas\u0131l toplan\u0131r<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"