{"id":25181,"date":"2022-12-08T13:57:05","date_gmt":"2022-12-08T08:27:05","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-programming-r-using-ifelse-with-multiple-conditions\/"},"modified":"2022-12-08T13:57:05","modified_gmt":"2022-12-08T08:27:05","slug":"solved-programming-r-using-ifelse-with-multiple-conditions","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-programming-r-using-ifelse-with-multiple-conditions\/","title":{"rendered":"[Solved] programming R using ifelse with multiple conditions"},"content":{"rendered":"<p> [ad_1]<br \/>\n<\/p>\n<div id=\"answer-53616719\" class=\"answer js-answer accepted-answer js-accepted-answer\" data-answerid=\"53616719\" data-parentid=\"53615890\" data-score=\"5\" data-position-on-page=\"1\" data-highest-scored=\"1\" data-question-has-accepted-highest-score=\"1\" itemprop=\"acceptedAnswer\" itemscope itemtype=\"https:\/\/schema.org\/Answer\">\n<div class=\"post-layout\">\n<div class=\"votecell post-layout--left\"><\/div>\n<div class=\"answercell post-layout--right\">\n<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>You can avoid the loop or the apply statement by vectorizing the <code>ifelse<\/code> statement.<\/p>\n<p>If you define <code>i<\/code> as the vector from 3:length(E$phone), you can then run the ifelse statement directly.<\/p>\n<pre><code>#test data\nphone&lt;-c(123,123,123,333,333,333,456,456,456,789,789,789,500,500,500)\ntime&lt;-c(\"2018-11-06\",\"2018-11-06\",\"2018-11-06\",\"2018-11-09\",\"2018-11-09\",\"2018-11-09\",\n        \"2018-11-07\",\"2018-11-07\",\"2018-11-07\",\"2018-11-05\",\"2018-11-05\", \"2018-11-05\", \n        \"2018-11-06\",\"2018-11-06\",\"2018-11-06\")\ntime&lt;-as.Date(time)\ntel&lt;-c(0,0,1,1,0,1,1,1,0,1,1,1,0,0,1)\nporad&lt;-c(1,2,3,1,2,3,1,2,3,1,2,3,1,2,3)\nE&lt;-data.frame(phone, time, tel, porad)\n\nE$de[1]=ifelse(E$phone[1]==E$phone[2] &amp; E$time[1]==E$time[2] &amp; E$porad[1]==2 &amp; E$tel[1]==1,1,0)\nE$de[2]=ifelse(E$phone[2]==E$phone[3] &amp; E$time[2]==E$time[3] &amp; E$porad[2]==3 &amp; E$tel[2]==1,1,0)\n\n#vectorized ifelse statement\ni&lt;-3:length(E$phone)\nE$de[i]&lt;-ifelse(E$phone[i]==E$phone[i-2] &amp; E$time[i]==E$time[i-2] &amp; E$porad[i]==3 &amp; E$tel[i]==1 &amp; E$tel[i-1]==0 &amp; E$tel[i-2]==0,1,0)\n<\/code><\/pre>\n<p>This should run approximately 1000x faster than the for loop.<\/p>\n<\/p><\/div>\n<div class=\"mt24\"><\/div>\n<\/div>\n<p>            <span class=\"d-none\" itemprop=\"commentCount\">0<\/span> <\/p><\/div>\n<\/div>\n<p>[ad_2]<\/p>\n<p>solved programming R using ifelse with multiple conditions <\/p>\n","protected":false},"excerpt":{"rendered":"<p>[ad_1] You can avoid the loop or the apply statement by vectorizing the ifelse statement. If you define i as the vector from 3:length(E$phone), you can then run the ifelse statement directly. #test data phone&lt;-c(123,123,123,333,333,333,456,456,456,789,789,789,500,500,500) time&lt;-c(&#8220;2018-11-06&#8243;,&#8221;2018-11-06&#8243;,&#8221;2018-11-06&#8243;,&#8221;2018-11-09&#8243;,&#8221;2018-11-09&#8243;,&#8221;2018-11-09&#8221;, &#8220;2018-11-07&#8243;,&#8221;2018-11-07&#8243;,&#8221;2018-11-07&#8243;,&#8221;2018-11-05&#8243;,&#8221;2018-11-05&#8221;, &#8220;2018-11-05&#8221;, &#8220;2018-11-06&#8243;,&#8221;2018-11-06&#8243;,&#8221;2018-11-06&#8221;) time&lt;-as.Date(time) tel&lt;-c(0,0,1,1,0,1,1,1,0,1,1,1,0,0,1) porad&lt;-c(1,2,3,1,2,3,1,2,3,1,2,3,1,2,3) E&lt;-data.frame(phone, time, tel, porad) E$de[1]=ifelse(E$phone[1]==E$phone[2] &amp; E$time[1]==E$time[2] &amp; E$porad[1]==2 &amp; E$tel[1]==1,1,0) E$de[2]=ifelse(E$phone[2]==E$phone[3] &amp; &#8230; <a title=\"[Solved] programming R using ifelse with multiple conditions\" class=\"read-more\" href=\"https:\/\/jassweb.com\/solved\/solved-programming-r-using-ifelse-with-multiple-conditions\/\" aria-label=\"More on [Solved] programming R using ifelse with multiple conditions\">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":[1544,321],"class_list":["post-25181","post","type-post","status-publish","format-standard","hentry","category-solved","tag-conditional-statements","tag-r"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.5 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>[Solved] programming R using ifelse with multiple conditions - 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-programming-r-using-ifelse-with-multiple-conditions\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] programming R using ifelse with multiple conditions - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] You can avoid the loop or the apply statement by vectorizing the ifelse statement. If you define i as the vector from 3:length(E$phone), you can then run the ifelse statement directly. #test data phone&lt;-c(123,123,123,333,333,333,456,456,456,789,789,789,500,500,500) time&lt;-c(&quot;2018-11-06&quot;,&quot;2018-11-06&quot;,&quot;2018-11-06&quot;,&quot;2018-11-09&quot;,&quot;2018-11-09&quot;,&quot;2018-11-09&quot;, &quot;2018-11-07&quot;,&quot;2018-11-07&quot;,&quot;2018-11-07&quot;,&quot;2018-11-05&quot;,&quot;2018-11-05&quot;, &quot;2018-11-05&quot;, &quot;2018-11-06&quot;,&quot;2018-11-06&quot;,&quot;2018-11-06&quot;) time&lt;-as.Date(time) tel&lt;-c(0,0,1,1,0,1,1,1,0,1,1,1,0,0,1) porad&lt;-c(1,2,3,1,2,3,1,2,3,1,2,3,1,2,3) E&lt;-data.frame(phone, time, tel, porad) E$de[1]=ifelse(E$phone[1]==E$phone[2] &amp; E$time[1]==E$time[2] &amp; E$porad[1]==2 &amp; E$tel[1]==1,1,0) E$de[2]=ifelse(E$phone[2]==E$phone[3] &amp; ... 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JassWeb","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/jassweb.com\/solved\/solved-programming-r-using-ifelse-with-multiple-conditions\/","og_locale":"en_US","og_type":"article","og_title":"[Solved] programming R using ifelse with multiple conditions - JassWeb","og_description":"[ad_1] You can avoid the loop or the apply statement by vectorizing the ifelse statement. If you define i as the vector from 3:length(E$phone), you can then run the ifelse statement directly. #test data phone&lt;-c(123,123,123,333,333,333,456,456,456,789,789,789,500,500,500) time&lt;-c(\"2018-11-06\",\"2018-11-06\",\"2018-11-06\",\"2018-11-09\",\"2018-11-09\",\"2018-11-09\", \"2018-11-07\",\"2018-11-07\",\"2018-11-07\",\"2018-11-05\",\"2018-11-05\", \"2018-11-05\", \"2018-11-06\",\"2018-11-06\",\"2018-11-06\") time&lt;-as.Date(time) tel&lt;-c(0,0,1,1,0,1,1,1,0,1,1,1,0,0,1) porad&lt;-c(1,2,3,1,2,3,1,2,3,1,2,3,1,2,3) E&lt;-data.frame(phone, time, tel, porad) E$de[1]=ifelse(E$phone[1]==E$phone[2] &amp; E$time[1]==E$time[2] &amp; E$porad[1]==2 &amp; E$tel[1]==1,1,0) E$de[2]=ifelse(E$phone[2]==E$phone[3] &amp; ... 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