2014-03-26 123 views
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我想根據三列之間的行方式比較在我的數據框中創建一個新列。一個示例數據幀:根據三個其他變量之間的比較創建新變量

df <- structure(list(year = c(1986L, 1987L, 1991L, 1991L, 1991L, 1991L, 1992L, 1992L, 1992L, 1992L, 1992L, 1992L, 1993L, 1993L, 1993L, 1993L, 1993L, 1993L, 1993L, 1993L, 1993L, 1993L, 1993L, 1993L, 1993L, 1993L, 1993L, 1993L, 1994L, 1994L, 1994L, 1994L, 1994L, 1994L, 1994L, 1994L, 1994L, 1994L, 1994L, 1994L, 1994L, 1994L, 1994L, 1994L, 1994L, 1994L, 1994L, 1994L, 1994L, 1994L), knmilocatie = structure(c(4L, 16L, 10L, 12L, 9L, 20L, 12L, 12L, 25L, 9L, 30L, 26L, 22L, 18L, 15L, 24L, 13L, 31L, 27L, 5L, 3L, 19L, 21L, 23L, 20L, 26L, 26L, 31L, 35L, 25L, 11L, 28L, 8L, 29L, 36L, 34L, 7L, 28L, 17L, 14L, 33L, 1L, 11L, 6L, 32L, 27L, 29L, 20L, 20L, 2L), .Label = c("Achterdiep", "Annen", "Appingedam", "Assen", "Bedum", "De Klip", "Delfzijl", "Eenrum", "Eleveld", "Emmen", "Garsthuizen", "Geelbroek", "Haren", "Hellum", "Hoogezand", "Hooghalen", "Kolham", "Langelo", "Loppersum", "Middelstum", "Nijenklooster", "Noordbroek", "Oldenzijl", "Overschild", "Roswinkel", "Slochteren", "Stedum", "Steendam", "t-Zandt", "Ten Boer", "Ten Post", "Uithuizermeeden", "Weiwerd", "Westerbroek", "Winneweer", "Zandeweer"), class = "factor"), baglocatie = structure(c(2L, 12L, 5L, 4L, 2L, 17L, 11L, 2L, 21L, 2L, 16L, 35L, 27L, 14L, 22L, 19L, 33L, 34L, 26L, 17L, 1L, 18L, 1L, 28L, 6L, 25L, 25L, 29L, 9L, 21L, 10L, 19L, 34L, 15L, 36L, 13L, 7L, 19L, 8L, 23L, 7L, 31L, 17L, 1L, 20L, 3L, 10L, 32L, 30L, 24L), .Label = c("Appingedam", "Assen", "Bedum", "Ekehaar", "Emmen", "Eppenhuizen", "Farmsum", "Froombosch", "Garrelsweer", "Garsthuizen", "Geelbroek", "Hooghalen", "Kolham", "Langelo", "Leermens", "Loppersum", "Middelstum", "Oosterwijtwerd", "Overschild", "Roodeschool", "Roswinkel", "Sappemeer", "Schildwolde", "Schipborg", "Slochteren", "Stedum", "Steendam", "t-Zandt", "Ten Post", "Toornwerd", "Tripscompagnie", "Warffum", "Westerbroek", "Wirdum", "Woudbloem", "Zandeweer"), class = "factor"), lllocatie = structure(c(3L, 13L, 5L, 10L, 4L, 32L, 10L, 10L, 22L, 4L, 36L, 37L, 31L, 15L, 23L, 20L, 34L, 8L, 24L, 35L, 19L, 19L, 2L, 29L, 26L, 25L, 25L, 30L, 8L, 22L, 9L, 20L, 19L, 16L, 38L, 12L, 6L, 27L, 7L, 11L, 17L, 33L, 14L, 2L, 21L, 18L, 9L, 28L, 32L, 1L), .Label = c("Annen", "Appingedam", "Assen", "Eleveld", "Emmen", "Farmsum", "Froombosch", "Garrelsweer", "Garsthuizen", "Geelbroek", "Hellum", "Hoogezand", "Hooghalen", "Huizinge", "Langelo", "Leermens", "Meedhuizen", "Onderdendam", "Oosterwijtwerd", "Overschild", "Roodeschool", "Roswinkel", "Sappemeer", "Sint Annen", "Slochteren", "Startenhuizen", "Steendam", "Stitswerd", "t-Zandt", "Ten Post", "Tjuchem", "Toornwerd", "Tripscompagnie", "Westerbroek", "Westerwijtwerd", "Winneweer", "Woudbloem", "Zandeweer"), class = "factor")), .Names = c("year", "knmilocatie", "baglocatie", "lllocatie"), class = "data.frame", row.names = c(NA, -50L)) 

當用於三列("knmilocatie", "baglocatie", "lllocatie")的值是在一排相同的(如在第一個三排的示例數據幀的情況下),新列應該得到的值該行代表只有一個唯一值的事實,即1。當所有三列具有不同的值時(請參閱第11行),新列的值應爲該行的值3。當比較中有兩個唯一值時,它應該得到值2

如何實現該建議?

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apply(df,1,function(u)length(unique(u)))? – droopy

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@droopy。也許你的意思是應用(df [-1],1,函數(u)長度(表格(u)))''。 –

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@droopy thanx,'apply(df [, - 1],1,function(u)length(unique(u)))'給了我正確的答案 – Jaap

回答

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兩種可能性:

apply(df[,-1], 1, function(u) length(unique(u))) # as you corrected it 
colSums(!apply(df[,-1], 1, duplicated)) 

第二種可能性可能會更快,如果你有一個大的數據集。 HTH