Question

我是大家，我删除了我的上一篇文章，以便重现我的问题。我正在处理数据帧的下一个a1（输入结构）：

structure(list(r04_numero_operacion = c("0050475725", "0050490602", 
"0050491033", "0050496386", "0050518985", "0050630090", "0050631615", 
"0060235906", "0060238732", "0060241333", "0060244391", "0060245813", 
"0060260056", "0060266356", "0800041441", "0800054041", "0800055382", 
"0800058554", "2020200062", "2020200073", "CAR1010001706000", 
"CAR1010001795000", "CAR1010001803000", "CAR1010001871000", "CAR1010001962000", 
"CAR1010002002000", "CAR1010002120000", "CAR1010002189000", "CAR1010002215000", 
"CAR1010002250000"), perdida3 = c(523.12, 265.43, 8371.66, 5242.13, 
4960.51, 8473.27, 3743.45, 1283.32, 2229.25, 8001.27, 8653.94, 
3670.13, 4536.02, 8216.55, 2481.36, 288.94, 1637.28, 4566.89, 
1573.63, 11217.92, 0, 0, 0, 0, 0, 0, 0, 0, 9633.9, 0), Saldo = c(1, 
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 288.94, 1637.28, 4566.89, 
1, 1, 481.59, 299.52, 258.13, 603.84, 231.61, 631.68, 220.6, 
210.54, 1, 1224.44), Bvencida = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 603.84, 0, 631.68, 
0, 0, 0, 0), Cvencida = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1224.44), 
    Dvencida = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), vencida = c(1, 
    1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 288.94, 1637.28, 
    4566.89, 1, 1, 0, 0, 0, 603.84, 0, 631.68, 0, 0, 1, 1224.44
    ), V1 = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 
    1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1)), .Names = c("r04_numero_operacion", 
"perdida3", "Saldo", "Bvencida", "Cvencida", "Dvencida", "vencida", 
"V1"), codepage = 1252L, row.names = c(NA, 30L), class = "data.frame")

a2数据框（dput结构）：

structure(list(r04_numero_operacion = c("0050475725", "0050490602", 
"0050491033", "0050496386", "0050518985", "0050630090", "0050631615", 
"0060235906", "0060238732", "0060241333", "0060244391", "0060245813", 
"0060260056", "0060266356", "0800041441", "0800054041", "0800055382", 
"0800058554", "2020200073", "CAR1010002002000", "CAR1010002189000", 
"CAR1010002215000", "CAR1010002250000", "CAR1010002264000", "CAR1010002297000", 
"CAR1010002401000", "CAR1010002412000", "CAR1010002436000", "CAR1010002529000", 
"CAR1010002709000"), perdida3 = c(523.12, 265.43, 8371.66, 5242.13, 
4960.51, 8473.27, 3743.45, 1283.32, 2229.25, 8001.27, 8653.94, 
3670.13, 4536.02, 8216.55, 2481.36, 288.94, 1637.28, 4566.89, 
11217.92, 0, 0, 9633.9, 0, 0, 0, 0, 0, 0, 0, 0), Saldo = c(1, 
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 288.94, 1637.28, 4566.89, 
1, 317.72, 210.54, 1, 868.93, 242.91, 298.78, 120.63, 255.01, 
357.68, 284.08, 308.83), Bvencida = c(0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 317.72, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0), Cvencida = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 868.93, 0, 0, 0, 0, 0, 0, 0), Dvencida = c(0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0), vencida = c(1, 1, 1, 1, 1, 1, 1, 1, 
1, 1, 1, 1, 1, 1, 1, 288.94, 1637.28, 4566.89, 1, 317.72, 0, 
1, 868.93, 0, 0, 0, 0, 0, 0, 0), V2 = c(2, 2, 2, 2, 2, 2, 2, 
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 
2, 2)), .Names = c("r04_numero_operacion", "perdida3", "Saldo", 
"Bvencida", "Cvencida", "Dvencida", "vencida", "V2"), class = "data.frame", row.names = c(NA, 
30L))

我的问题是当我使用merge()和match()函数时。 merge()比通过普通变量添加新变量的match()更有用，但是当我使用merge()时，我得到的结果与match()不同。首先，我使用merge()与a2和a1一起使用下一个代码创建DF：

DF=merge(a2,a1,all.x=TRUE)

它将V1变量从a1添加到DF，我得到了DF$V1的摘要：

Min. 1st Qu.  Median    Mean 3rd Qu.    Max.    NA's 
  1       1       1       1       1       1       9

创建名为a2的{{1}}副本后，我与DF匹配，使用此代码将r04_numero_operacion变量从V1添加到{ {1}}：

a1

它向a2添加了a2$V1<-a1[match(a2$r04_numero_operacion,a1$r04_numero_operacion),"V1"]，但结果与`V1方式不同。我在DF解决方案中获得了merge()的摘要：

DF$V1

我的问题是我希望与match()做同样的事情，但由于此功能而使用Min. 1st Qu. Median Mean 3rd Qu. Max. NA's 1 1 1 1 1 1 7功能比match()更强大。谢谢你的帮助。

Answer 1

在使用match(a2$r04_numero_operacion,a1$r04_numero_operacion)时，a2 $ r04_numero_operacion值与a1中的相应列匹配，而在使用merge(a2,a1,all.x=TRUE) a1时，所有匹配列都与a2中的匹配列名匹配。如果您只匹配第一列，则NA计数匹配：

summary( merge(a2,a1,by=1,all.x=TRUE)$V1 )
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.    NA's 
      1       1       1       1       1       1       7

R中合并和匹配函数之间的差异

1 个答案: