2012-03-18 34 views
3

[R初學者用什麼似乎是一個非常簡單的問題: 我有一些電子郵件的日誌,我已經在格式讀入R:R:轉換電子郵件地址爲唯一整數

>log1 
    Date  Time   From     To 
1 2000-01-01 00:00:00 [email protected]   [email protected] 
2 2000-01-02 01:00:00 carolyn @mail.com  [email protected] 
3 2000-01-03 02:00:00 [email protected]   [email protected] 
4 2000-01-04 03:00:00 chris @mail.com   [email protected] 
5 2000-01-05 04:00:00 [email protected]   [email protected] 
6 2000-01-06 05:00:00 [email protected]   [email protected] 

我需要要將log1 $ From和log1 $ To更改爲全局唯一數字標識符,以便稍後在其他日誌中讀取任何給定電子郵件地址時將收到與先前日誌相同的標識符。

我曾嘗試:

id <- as.numeric(as.character(log1[,3]))) 
id<-as.numeric(levels(log1[,3]))) 
id <- charToRaw(log1[,4]), base=16) 

會某種靈魂請幫我 - 謝謝!

道歉或許應該已經包括此:

Date=c("01/01/2000" ,"02/01/2000" ,"03/01/2000", "04/01/2000" ,"05/01/2000" ,"06/01/2000","07/01/2000","08/01/2000", 
    "09/01/2000","10/01/2000","11/01/2000", "12/01/2000" ,"13/01/2000", "14/01/2000", "15/01/2000","16/01/2000" 
    ,"17/01/2000","18/01/2000","19/01/2000","20/01/2000","01/01/2000","02/01/2000") 
    Time=c("00:00:00","01:00:00","02:00:00", "03:00:00" ,"04:00:00" ,"05:00:00", "06:00:00" ,"07:00:00", "08:00:00", "09:00:00" ,"10:00:00", 
    "11:00:00", "12:00:00","13:00:00", "14:00:00","15:00:00","16:00:00","17:00:00","18:00:00","19:00:00","00:00:00" ,"00:00:00") 
    From=c("[email protected]","[email protected]","[email protected]","[email protected]","[email protected]","[email protected]", 
    "[email protected]","[email protected]","[email protected]","[email protected]","[email protected]","[email protected]", 
    "[email protected]","[email protected]","[email protected]","[email protected]","[email protected]","[email protected]", 
    "[email protected]","[email protected]","[email protected]","[email protected]") 
    To=c("[email protected]","[email protected]","[email protected]","[email protected]","[email protected]","[email protected]","[email protected]", 
    "[email protected]","[email protected]","[email protected]","[email protected]","[email protected]","[email protected]","[email protected]", 
    "[email protected]","[email protected]","[email protected]","[email protected]","[email protected]","[email protected]","[email protected]","[email protected]") 
    log<-data.frame(Date=Date,Time=Time,From=From,To=To) 

在嘗試使用MD5生成全局唯一標識符:注意[email protected]標識符是如何內ID_to正確的比賽,但不在範圍在水平/因子方法ID_from

ID_to<-data.frame() 
    ID_from<-data.frame() 

    for (i in 1:nrow(log)){ 
    to<-as.numeric(paste('0x', substr(rep(hmac('secret',log[i,4], algo='md5'), 2), c(1, 9, 17, 25), c(8, 16, 24, 32)),sep="")) 
    (ID_to<-rbind(ID_to,to)) 

    from<-as.numeric(paste('0x', substr(rep(hmac('secret',log[i,3], algo='md5'), 2), c(1, 9, 17, 25),c(8, 16, 24, 32)),sep="")) 
    (ID_from<-rbind(ID_from,from)) 

    } 

    ID_to[,3]<-paste(ID_to[,1],ID_to[,2], sep="") 
    ID_from[,3]<-paste(ID_from[,1],ID_from[,2], sep="") 

    edgelist<-data.frame(ID_from[,3],log[,3],ID_to[,3],log[,4],log[,1],log[,2]) 
    print(edgelist) 
    ID_from...3.     log...3.   ID_to...3.   log...4. log...1. log...2. 
    27488842661591306920  [email protected] 18727221862165338513 [email protected] 01/01/2000 00:00:00 
    38124472891255273775 [email protected] 1251903296725454474  [email protected] 02/01/2000 01:00:00 
    29070047663451376630  [email protected] 17074276751156451031  [email protected] 03/01/2000 02:00:00 
    8261398433828474582 [email protected] 1563683670909194033  [email protected] 04/01/2000 03:00:00 
    18727221862165338513 [email protected] 26735368323826533112  [email protected] 05/01/2000 04:00:00 
    5680838251168988404  [email protected] 2923605896229594830  [email protected] 06/01/2000 05:00:00 
    2351312285811012730  [email protected] 17171333544033270402  [email protected] 07/01/2000 06:00:00 
    328278708432069254  [email protected] 33840664403556851587  [email protected] 08/01/2000 07:00:00 
    1127901879852039037 [email protected] 1973548136161209824  [email protected] 09/01/2000 08:00:00 
    7349515121496417787 [email protected] 5680838251168988404  [email protected] 10/01/2000 09:00:00 
    27488842661591306920  [email protected] 328278708432069254  [email protected] 11/01/2000 10:00:00 
    38124472891255273775 [email protected] 1127901879852039037  [email protected] 12/01/2000 11:00:00 
    29070047663451376630  [email protected] 27488842661591306920  [email protected] 13/01/2000 12:00:00 
    8261398433828474582 [email protected] 38124472891255273775  [email protected] 14/01/2000 13:00:00 
    18727221862165338513 [email protected] 29070047663451376630  [email protected] 15/01/2000 14:00:00 
    5680838251168988404  [email protected] 8261398433828474582  [email protected] 16/01/2000 15:00:00 
    2351312285811012730  [email protected] 2351312285811012730  [email protected] 17/01/2000 16:00:00 
    328278708432069254  [email protected] 7349515121496417787  [email protected] 18/01/2000 17:00:00 
    1127901879852039037 [email protected] 41762759923562968495  [email protected] 19/01/2000 18:00:00 
    7349515121496417787 [email protected] 24894056753582090007  [email protected] 20/01/2000 19:00:00 
    27488842661591306920  [email protected] 18727221862165338513 [email protected] 01/01/2000 00:00:00 
    27488842661591306920  [email protected] 18727221862165338513 [email protected] 02/01/2000 00:00:00 

嘗試:

獲得一個錯誤:

log <- union(levels(log[,3]), levels(log[,4])) 
>Error in emails[, 3] : incorrect number of dimensions 
+0

不太瞭解R,但是從你提到的內容來看,你正在尋找From和To電子郵件地址組合的唯一標識符。你可以嘗試爲它們的連接創建一個散列。 R似乎有一些散列函數,所以你可以嘗試一下。 – Gangadhar 2012-03-18 15:09:43

+0

感謝您的輸入傢伙,當然有一個比實現校驗和或hashmap更簡單的解決方案?! – 2012-03-18 15:41:18

+0

只要您爲每個輸入獲取唯一標識符,就可以使用任何算法(md5,sha,crc,..)。 – blejzz 2012-03-18 16:07:51

回答

1

您需要爲日誌中的每封電子郵件創建唯一的ID。一種方法是計算每封電子郵件的crc校驗和,並將其用作標識符,但數字會很長。或者你可以在R中實現一個HashMap,並將該電子郵件作爲HashMap的關鍵字。

2

您可以使用MD5生成全局唯一標識符,因爲它的衝突概率非常低,但由於其輸出爲128位,因此需要一些數字來表示它(32位R中有4個整數,兩個整數在64位R中)。不過,這應該很容易處理使用短數字向量。

這裏是你如何生成的電子郵件地址這樣的四個整數向量(或任何其他字符串爲此事):

library(digest) 
email <- '[email protected]' 
as.numeric(paste('0x', substr(rep(hmac('secret56f8a7', email, algo='md5'), 4), c(1, 9, 17, 25), c(8, 16, 24, 32)), sep='')) 

你可以使用algo='crc32'和只獲得一個整數,但是這個ISN我們不建議這樣做,因爲碰撞比CRC更有可能。

+0

感謝您的輸入 - 不幸的是,這種解決方案似乎並沒有爲我工作。看看上面的輸出,我(非常)可能會做出一些單子。謝謝 – 2012-03-19 00:04:08

1

我認爲這會做你想要什麼,這是有效的,你可以只使用基礎包做...

步驟:

1.Convert的兩列因素

2.以完全相同的方式聯合因素級別,以便每個電子郵件在因素級別中都有唯一的ID。

3.將每列中的條目更改爲與其因子級別對應的數字。因此,我們可以通過簡單地在兩欄中查找「1」來識別「[email protected]」發送和接收電子郵件的時間。

log1$From <- as.factor(log1$From) 
log1$To <- as.factor(log1$To) 
emails <- union(levels(log1$From), levels(log1$To)) 
levels(log1$From) <- emails 
levels(log1$To) <- emails 
log1$From <- as.numeric(log1$From) 
log1$To <- as.numeric(log1$To) 

如我在這裏所做的那樣,保留原始電子郵件地址的記錄可能是一個好主意。然後,如果你有興趣,比如發送哪些郵件[email protected]

log1[log1$From == which(emails == "[email protected]"), ] 

應該這樣做!你可以編寫一個程序,使它看起來更清潔...

+0

,感謝您的輸入 - 不幸的是,我遇到了這個解決方案的錯誤。 – 2012-03-19 00:23:01

+0

不幸的是,這並不能確保在閱讀不同的日誌時,同一個電子郵件地址將獲得與OP所需的相同的數字ID。 – 2012-03-19 09:39:23

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