RE: blocks, noreplacenorepeat

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blocks, noreplacenorepeat

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jo700	jo700 posted 9 Years Ago Topic Details Share Topic
Group: Forum Members Posts: 17, Visits: 84	Hey everyone, I've got three blocks. In every block 9 different pictures should be shown in a random order. Could anyone help me to make it work? Unfortunately I did not find it under Inquisit help and tried different ways to handle it. But somehow I don't get it. Hope you are can help me. Thanks a lot! <block block4A> / trials = [1-9=noreplacenorepeat (1=Bild1A; 2=Bild2A; 3=Bild3A; 4=Bild4A; 5=Bild5A; 6=Bild6A; 7=Bild7A; 8=Bild8A; 9=Bild9A)] </block> <block block4B> / trials = [1-9=noreplacenorepeat (1=Bild1B; 2=Bild2B; 3=Bild3B; 4=Bild4B; 5=Bild5B; 6=Bild6B; 7=Bild7B; 8=Bild8B; 9=Bild9B)] </block> <block block4C> / trials = [1-9=noreplacenorepeat (1=Bild1c; 2=Bild2C; 3=Bild3C; 4=Bild4C; 5=Bild5C; 6=Bild6C; 7=Bild7C; 8=Bild8C; 9=Bild9C)] </block>
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Dave	Dave posted 9 Years Ago ANSWER Post Details Share Post
Group: Administrators Posts: 13K, Visits: 105K	Supposing that "Bild1A" etc. are <trial> elements, <block block4A> / trials = [1-9=noreplacenorepeat (1=Bild1A; 2=Bild2A; 3=Bild3A; 4=Bild4A; 5=Bild5A; 6=Bild6A; 7=Bild7A; 8=Bild8A; 9=Bild9A)] </block> needs to read <block block4A> / trials = [1-9=noreplacet (Bild1A, Bild2A, Bild3A, Bild4A, Bild5A, Bild6A, Bild7A, Bild8A, Bild9A)] </block> There is no need to noreplacenorepeat, since you are taking 9 samples from 9 elements -- there is nothing that could possibly repeat.
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jo700	jo700 posted 9 Years Ago ANSWER Post Details Share Post
Group: Forum Members Posts: 17, Visits: 84	Thanks a lot! :-)
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