優化R中的循環

我想優化R中的以下代碼。此循環需要很長時間才能運行。我不知道是否有人可以幫我優化這段代碼，因爲它需要很長時間才能運行？謝謝大家！優化R中的循環

SIN_FM5：是一個包含約300,000行和7列的數據庫。

# Make Combination 

SIN_FM5$Combination=(SIN_FM5$SINISTRE) 
Count.Comb=data.frame(table(SIN_FM5$Combination)) 


# Calculate number of combinations 
Total.Comb=nrow(Count.Comb) 

# Loop through all combinations and calculate statistics 

Statistics=array(0,dim=c(Total.Comb,5)) 

for (i in 1:Total.Comb) { 

    Subset=subset(SIN_FM5, SIN_FM5$Combination==Count.Comb[i,1]) 
    Statistics[(i),]=c(Count.Comb[i,1],mean(Subset$MONTANT_PAIEMENT), 
    median(Subset$MONTANT_PAIEMENT),min(Subset$MONTANT_PAIEMENT), 
    max(Subset$MONTANT_PAIEMENT)) 

} 
resultatN=cbind(Count.Comb,Statistics) 

dput(head(SIN_FM5))

控制檯〜/

"TRSP-5194", "TRSP-5197", "TRSP-5201", "TRSP-5202", "TRSP-5204", 
"TRSP-5205", "TRSP-5207", "TRSP-5212", "TRSP-5214", "TRSP-5215", 
"TRSP-5218", "TRSP-5222", "TRSP-5230", "TRSP-5238", "TRSP-5243", 
"TRSP-5247", "TRSP-5248", "TRSP-5253", "TRSP-5254", "TRSP-5255", 
"TRSP-5257", "TRSP-5259", "TRSP-5262", "TRSP-5263", "TRSP-5266", 
"TRSP-5267", "TRSP-5268", "TRSP-5270", "TRSP-5271", "TRSP-5274", 
"TRSP-5277", "TRSP-5279", "TRSP-5281", "TRSP-5283", "TRSP-5288", 
"TRSP-5289", "TRSP-5293", "TRSP-5296", "TRSP-5299", "TRSP-5301", 
"TRSP-5303", "TRSP-5304", "TRSP-5306", "TRSP-5308", "TRSP-5310", 
"TRSP-5311", "TRSP-5312", "TRSP-5313", "TRSP-5335", "TRSP-5343", 
"TRSP-5348", "TRSP-5352", "TRSP-5357", "TRSP-5363", "TRSP-5366", 
"TRSP-5372", "TRSP-5373", "TRSP-5384", "TRSP-5386", "TRSP-5388", 
"TRSP-5391", "TRSP-5392", "TRSP-5428", "TRSP-5436", "VANBILSENYolanda", 
"VanLierop", "VirgaJesseZiekenhuis", "WanetGeorges", "WANETThierry", 
"WILLEMSMichel", "WUESTENBERGHSAlain", "X01", "X01CR", "X02CR", 
"X03CR", "X04CR", "X05CR", "X06CR", "X07CR", "Y01", "Y01CR", 
"Y02", "ZOPO-5344"), class = "factor"), Combination = c(73010009L, 
73010009L, 73010014L, 73010014L, 73010014L, 73010014L)), .Names = c("SINISTRE", 
"victimeid", "Nature.Injury", "LocationL", "DurationITT", "Code.Nace2008", 
"POLICE", "TYPE_DE_PAIEMENT", "MODE_DE_PAIEMENT", "CODE_NATURE_DE_PAIEMENT", 
"MONTANT_PAIEMENT", "BENEFICIARE", "Combination"), row.names = c(NA, 
6L), class = "data.frame")

來源

2012-07-01 Yasmine Nouri

類= 「因子」），組合= C（73010009L， 73010009L，73010014L，73010014L，73010014L，73010014L）），.Names = C（「SINISTRE」，「victimeid 「，」自然傷害「，」LocationL「，」DurationITT「，」Code.Nace2008「，」POLICE「，」TYPE_DE_PAIEMENT「，」MODE_DE_PAIEMENT「，」CODE_NATURE_DE_PAIEMENT「，」MONTANT_PAIEMENT「，」BENEFICIARE「「），row.names = c（NA， 6L），class =」data.frame「） –

它還給出了我的數據幀變量」BENEFICIARE「的所有值SIN_FM5 –

您可以將* complete *的輸出在你的問題中輸入（頭（SIN_FM5））'。 – mnel

SIN_FM5 <- data.frame(Combination = sample(1:10, 100, repl=TRUE), MONTANT_PAIEMENT=rnorm(100)) 
bySIN <- by(SIN_FM5, list(SIN_FM5[['Combination']]), FUN= function(subd) { 
      data.frame(counts = nrow(subd), 
       meanMont = mean(subd$MONTANT_PAIEMENT), 
       medMont = median(subd$MONTANT_PAIEMENT), 
       minMont = min(subd$MONTANT_PAIEMENT), 
       maxMont = max(subd$MONTANT_PAIEMENT)) }) 
> sapply(bySIN, as.vector) 
     1   2   3   4   5   6   7   8   
counts 11  7   14   9   11   16   10   8   
meanMont 0.3499753 -0.188964 0.1740817 -0.1505312 -0.6335896 -0.1434513 -0.2148642 -0.2978299 
medMont 0.4381513 -0.06965143 0.05762425 -0.2247187 -0.7682626 -0.2288606 -0.1467318 -0.3315809 
minMont -1.122418 -0.9638749 -1.634259 -1.336908 -2.068224 -1.974108 -2.15415 -1.295045 
maxMont 1.50662 0.1653189 1.215114 1.243138 0.4643551 1.29805 1.154282 0.7097163 
     9   10  
counts 4   10  
meanMont 0.2575141 0.146613 
medMont 0.07692888 0.1047567 
minMont -0.534418 -1.006938 
maxMont 1.410617 1.4973

這裏有一個data.table解決方案。可能要快得多：

require(data.table) 
dtb <- data.table(SIN_FM5) 
setkey(dtb, "Combination") 

dtb[ , list(counts=length(MONTANT_PAIEMENT), 
       meanMont = mean(MONTANT_PAIEMENT), 
       medMont = median(MONTANT_PAIEMENT), 
       minMont = min(MONTANT_PAIEMENT), 
       maxMont = max(MONTANT_PAIEMENT)), by="Combination"] 
#----------------------------------------------- 
     Combination counts meanMont  medMont minMont maxMont 
[1,]   1  11 0.3499753 0.43815133 -1.1224179 1.5066198 
[2,]   2  7 -0.1889640 -0.06965143 -0.9638749 0.1653189 
[3,]   3  14 0.1740817 0.05762425 -1.6342586 1.2151136 
[4,]   4  9 -0.1505312 -0.22471868 -1.3369085 1.2431378 
[5,]   5  11 -0.6335896 -0.76826264 -2.0682244 0.4643551 
[6,]   6  16 -0.1434513 -0.22886060 -1.9741083 1.2980502 
[7,]   7  10 -0.2148642 -0.14673175 -2.1541500 1.1542819 
[8,]   8  8 -0.2978299 -0.33158086 -1.2950452 0.7097163 
[9,]   9  4 0.2575141 0.07692888 -0.5344180 1.4106168 
[10,]   10  10 0.1466130 0.10475674 -1.0069382 1.4972998

來源

2012-07-01 23:48:38

它向我發送此消息：aggregate.data.frame中的錯誤（SIN_FM5，by = SIN_FM5 $組合，FUN = function（subd）{' 'by'必須是列表 –

我用'1基於'by' –

這似乎工作，但是否有可能爲每個SIN_FM5 $ SINISTRE（這是每個受害者的ID號碼）的「montant_paiement」的統計計算？ –

優化R中的循環

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