{"id":2038,"date":"2023-07-23T23:50:20","date_gmt":"2023-07-23T23:50:20","guid":{"rendered":"https:\/\/statorials.org\/tr\/r-kosulunun-uzunlugu-1dir-yalnizca-ilk-eleman-kullanilacaktir\/"},"modified":"2023-07-23T23:50:20","modified_gmt":"2023-07-23T23:50:20","slug":"r-kosulunun-uzunlugu-1dir-yalnizca-ilk-eleman-kullanilacaktir","status":"publish","type":"post","link":"https:\/\/statorials.org\/tr\/r-kosulunun-uzunlugu-1dir-yalnizca-ilk-eleman-kullanilacaktir\/","title":{"rendered":"R'de nas\u0131l d\u00fczeltilir: ko\u015fulun uzunlu\u011fu > 1'dir ve yaln\u0131zca ilk \u00f6\u011fe kullan\u0131lacakt\u0131r"},"content":{"rendered":"

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R’de kar\u015f\u0131la\u015fabilece\u011finiz bir hata:<\/span><\/p>\n

 Warning message:\nIn if (x > 1) {:\n  the condition has length > 1 and only the first element will be used \n<\/strong><\/pre>\n
 Bu hata, bir ko\u015fulu kontrol etmek i\u00e7in if()<\/strong> i\u015flevini kullanmaya \u00e7al\u0131\u015ft\u0131\u011f\u0131n\u0131zda ancak if()<\/strong> i\u015flevine tek tek \u00f6\u011feler yerine bir vekt\u00f6r iletti\u011finizde ortaya \u00e7\u0131kar.<\/span><\/p>\n
 Bu e\u011fitimde bu hatan\u0131n nas\u0131l d\u00fczeltilece\u011fi tam olarak a\u00e7\u0131klanmaktad\u0131r.<\/span><\/p>\n
 Hata nas\u0131l yeniden olu\u015fturulur?<\/strong><\/h3>\n
 R’de a\u015fa\u011f\u0131daki vekt\u00f6re sahip oldu\u011fumuzu varsayal\u0131m:<\/span><\/p>\n
 #define data\nx <- c(2, 3, 1, 1, 5, 7)<\/span>\n<\/span><\/strong><\/span><\/pre>\n
 \u015eimdi x vekt\u00f6r\u00fcn\u00fcn her de\u011ferinin 1’den b\u00fcy\u00fck olup olmad\u0131\u011f\u0131n\u0131 kontrol etmek i\u00e7in if()<\/strong> i\u015flevini kullanmaya \u00e7al\u0131\u015ft\u0131\u011f\u0131m\u0131z\u0131 ve ard\u0131ndan bu de\u011ferleri 2 ile \u00e7arpt\u0131\u011f\u0131m\u0131z\u0131 varsayal\u0131m:<\/span><\/p>\n
 #if value in vector x is greater than 1, multiply it by 2\nif (x>1) {\n  x*2\n}\n\n<\/span><\/span>Warning message:\nIn if (x > 1) {:\n  the condition has length > 1 and only the first element will be used\n<\/strong><\/span><\/pre>\n
 if()<\/strong> ifadesine bir vekt\u00f6r aktard\u0131\u011f\u0131m\u0131z i\u00e7in bir uyar\u0131 mesaj\u0131 al\u0131yoruz.<\/span><\/p>\n
 Bir if()<\/strong> ifadesi bir kerede bir vekt\u00f6r\u00fcn yaln\u0131zca bir eleman\u0131n\u0131 kontrol edebilir, ancak bu kodu kullanarak vekt\u00f6r\u00fcn her bir eleman\u0131n\u0131 bir kerede kontrol etmeye \u00e7al\u0131\u015ft\u0131k.<\/span><\/p>\n
 Hata nas\u0131l d\u00fczeltilir?<\/strong><\/span><\/h3>\n
 Bu hatay\u0131 d\u00fczeltmenin en kolay yolu bunun yerine ifelse()<\/strong> i\u015flevini kullanmakt\u0131r:<\/span><\/p>\n
 #if value in vector x is greater than 1, multiply it by 2\n<\/span>ifelse(x>1, x*2, x)\n\n[1] 4 6 1 1 10 14\n<\/strong><\/span><\/pre>\n
 Varsay\u0131lan olarak ifelse()<\/strong> i\u015flevi bir vekt\u00f6r\u00fcn her \u00f6\u011fesini tek tek kontrol eder. Bu, daha \u00f6nce kar\u015f\u0131la\u015ft\u0131\u011f\u0131m\u0131z hatadan ka\u00e7\u0131nmam\u0131z\u0131 sa\u011flar.<\/span><\/p>\n
 ifelse()<\/strong> fonksiyonunun \u00fcretti\u011fi \u00e7\u0131kt\u0131 de\u011ferlerini \u015fu \u015fekilde \u00fcretmektedir:<\/span><\/p>\n