How to do vlookup and fill down (like in Excel) in

问题:

I have a dataset about 105000 rows and 30 columns. I have a categorical variable that I would like to assign it to a number. In Excel, I would probably do something with VLOOKUP and fill.

How would I go about doing the same thing in R?

Essentially, what I have is a HouseType variable, and I need to calculate the HouseTypeNo. Here are some sample data:

HouseType HouseTypeNo
Semi            1
Single          2
Row             3
Single          2
Apartment       4
Apartment       4
Row             3

回答1:

If I understand your question correctly, here are four methods to do the equivalent of Excel\'s VLOOKUP and fill down using R:

# load sample data from Q
hous <- read.table(header = TRUE, 
                   stringsAsFactors = FALSE, 
text=\"HouseType HouseTypeNo
Semi            1
Single          2
Row             3
Single          2
Apartment       4
Apartment       4
Row             3\")

# create a toy large table with a \'HouseType\' column 
# but no \'HouseTypeNo\' column (yet)
largetable <- data.frame(HouseType = as.character(sample(unique(hous$HouseType), 1000, replace = TRUE)), stringsAsFactors = FALSE)

# create a lookup table to get the numbers to fill
# the large table
lookup <- unique(hous)
  HouseType HouseTypeNo
1      Semi           1
2    Single           2
3       Row           3
5 Apartment           4

Here are four methods to fill the HouseTypeNo in the largetable using the values in the lookup table:

First with merge in base:

# 1. using base 
base1 <- (merge(lookup, largetable, by = \'HouseType\'))

A second method with named vectors in base:

# 2. using base and a named vector
housenames <- as.numeric(1:length(unique(hous$HouseType)))
names(housenames) <- unique(hous$HouseType)

base2 <- data.frame(HouseType = largetable$HouseType,
                    HouseTypeNo = (housenames[largetable$HouseType]))

Third, using the plyr package:

# 3. using the plyr package
library(plyr)
plyr1 <- join(largetable, lookup, by = \"HouseType\")

Fourth, using the sqldf package

# 4. using the sqldf package
library(sqldf)
sqldf1 <- sqldf(\"SELECT largetable.HouseType, lookup.HouseTypeNo
FROM largetable
INNER JOIN lookup
ON largetable.HouseType = lookup.HouseType\")

If it\'s possible that some house types in largetable do not exist in lookup then a left join would be used:

sqldf(\"select * from largetable left join lookup using (HouseType)\")

Corresponding changes to the other solutions would be needed too.

Is that what you wanted to do? Let me know which method you like and I\'ll add commentary.

回答2:

I think you can also use match():

largetable$HouseTypeNo <- with(lookup,
                     HouseTypeNo[match(largetable$HouseType,
                                       HouseType)])

This still works if I scramble the order of lookup.

回答3:

I also like using qdapTools::lookup or shorthand binary operator %l%. It works identically to an Excel vlookup, but it accepts name arguments opposed to column numbers

## Replicate Ben\'s data:
hous <- structure(list(HouseType = c(\"Semi\", \"Single\", \"Row\", \"Single\", 
    \"Apartment\", \"Apartment\", \"Row\"), HouseTypeNo = c(1L, 2L, 3L, 
    2L, 4L, 4L, 3L)), .Names = c(\"HouseType\", \"HouseTypeNo\"), 
    class = \"data.frame\", row.names = c(NA, -7L))


largetable <- data.frame(HouseType = as.character(sample(unique(hous$HouseType), 
    1000, replace = TRUE)), stringsAsFactors = FALSE)


## It\'s this simple:
library(qdapTools)
largetable[, 1] %l% hous

回答4:

Solution #2 of @Ben\'s answer is not reproducible in other more generic examples. It happens to give the correct lookup in the example because the unique HouseType in houses appear in increasing order. Try this:

hous <- read.table(header = TRUE,   stringsAsFactors = FALSE,   text=\"HouseType HouseTypeNo
  Semi            1
  ECIIsHome       17
  Single          2
  Row             3
  Single          2
  Apartment       4
  Apartment       4
  Row             3\")

largetable <- data.frame(HouseType = as.character(sample(unique(hous$HouseType), 1000, replace = TRUE)), stringsAsFactors = FALSE)
lookup <- unique(hous)

Bens solution#2 gives

housenames <- as.numeric(1:length(unique(hous$HouseType)))
names(housenames) <- unique(hous$HouseType)
base2 <- data.frame(HouseType = largetable$HouseType,
                    HouseTypeNo = (housenames[largetable$HouseType]))

which when

unique(base2$HouseTypeNo[ base2$HouseType==\"ECIIsHome\" ])
[1] 2

when the correct answer is 17 from the lookup table

The correct way to do it is

 hous <- read.table(header = TRUE,   stringsAsFactors = FALSE,   text=\"HouseType HouseTypeNo
      Semi            1
      ECIIsHome       17
      Single          2
      Row             3
      Single          2
      Apartment       4
      Apartment       4
      Row             3\")

largetable <- data.frame(HouseType = as.character(sample(unique(hous$HouseType), 1000, replace = TRUE)), stringsAsFactors = FALSE)

housenames <- tapply(hous$HouseTypeNo, hous$HouseType, unique)
base2 <- data.frame(HouseType = largetable$HouseType,
  HouseTypeNo = (housenames[largetable$HouseType]))

Now the lookups are performed correctly

unique(base2$HouseTypeNo[ base2$HouseType==\"ECIIsHome\" ])
ECIIsHome 
       17

I tried to edit Bens answer but it gets rejected for reasons I cannot understand.

回答5:

Starting with:

houses <- read.table(text=\"Semi            1
Single          2
Row             3
Single          2
Apartment       4
Apartment       4
Row             3\",col.names=c(\"HouseType\",\"HouseTypeNo\"))

... you can use

as.numeric(factor(houses$HouseType))

... to give a unique number for each house type. You can see the result here:

> houses2 <- data.frame(houses,as.numeric(factor(houses$HouseType)))
> houses2
  HouseType HouseTypeNo as.numeric.factor.houses.HouseType..
1      Semi           1                                    3
2    Single           2                                    4
3       Row           3                                    2
4    Single           2                                    4
5 Apartment           4                                    1
6 Apartment           4                                    1
7       Row           3                                    2

... so you end up with different numbers on the rows (because the factors are ordered alphabetically) but the same pattern.

(EDIT: the remaining text in this answer is actually redundant. It occurred to me to check and it turned out that read.table() had already made houses$HouseType into a factor when it was read into the dataframe in the first place).

However, you may well be better just to convert HouseType to a factor, which would give you all the same benefits as HouseTypeNo, but would be easier to interpret because the house types are named rather than numbered, e.g.:

> houses3 <- houses
> houses3$HouseType <- factor(houses3$HouseType)
> houses3
  HouseType HouseTypeNo
1      Semi           1
2    Single           2
3       Row           3
4    Single           2
5 Apartment           4
6 Apartment           4
7       Row           3
> levels(houses3$HouseType)
[1] \"Apartment\" \"Row\"       \"Semi\"      \"Single\"  

回答6:

The poster didn\'t ask about looking up values if exact=FALSE, but I\'m adding this as an answer for my own reference and possibly others.

If you\'re looking up categorical values, use the other answers.

Excel\'s vlookup also allows you to match match approximately for numeric values with the 4th argument(1) match=TRUE. I think of match=TRUE like looking up values on a thermometer. The default value is FALSE, which is perfect for categorical values.

If you want to match approximately (perform a lookup), R has a function called findInterval, which (as the name implies) will find the interval / bin that contains your continuous numeric value.

However, let\'s say that you want to findInterval for several values. You could write a loop or use an apply function. However, I\'ve found it more efficient to take a DIY vectorized approach.

Let\'s say that you have a grid of values indexed by x and y:

grid <- list(x = c(-87.727, -87.723, -87.719, -87.715, -87.711), 
             y = c(41.836, 41.839, 41.843, 41.847, 41.851), 
             z = (matrix(data = c(-3.428, -3.722, -3.061, -2.554, -2.362, 
                                  -3.034, -3.925, -3.639, -3.357, -3.283, 
                                  -0.152, -1.688, -2.765, -3.084, -2.742, 
                                   1.973,  1.193, -0.354, -1.682, -1.803, 
                                   0.998,  2.863,  3.224,  1.541, -0.044), 
                         nrow = 5, ncol = 5)))

and you have some values you want to look up by x and y:

df <- data.frame(x = c(-87.723, -87.712, -87.726, -87.719, -87.722, -87.722), 
                 y = c(41.84, 41.842, 41.844, 41.849, 41.838, 41.842), 
                 id = c(\"a\", \"b\", \"c\", \"d\", \"e\", \"f\")

Here is the example visualized:

contour(grid)
points(df$x, df$y, pch=df$id, col=\"blue\", cex=1.2)

You can find the x intervals and y intervals with this type of formula:

xrng <- range(grid$x)
xbins <- length(grid$x) -1
yrng <- range(grid$y)
ybins <- length(grid$y) -1
df$ix <- trunc( (df$x - min(xrng)) / diff(xrng) * (xbins)) + 1
df$iy <- trunc( (df$y - min(yrng)) / diff(yrng) * (ybins)) + 1

You could take it one step further and perform a (simplistic) interpolation on the z values in grid like this:

df$z <- with(df, (grid$z[cbind(ix, iy)] + 
                      grid$z[cbind(ix + 1, iy)] +
                      grid$z[cbind(ix, iy + 1)] + 
                      grid$z[cbind(ix + 1, iy + 1)]) / 4)

Which gives you these values:

contour(grid, xlim = range(c(grid$x, df$x)), ylim = range(c(grid$y, df$y)))
points(df$x, df$y, pch=df$id, col=\"blue\", cex=1.2)
text(df$x + .001, df$y, lab=round(df$z, 2), col=\"blue\", cex=1)

df
#         x      y id ix iy        z
# 1 -87.723 41.840  a  2  2 -3.00425
# 2 -87.712 41.842  b  4  2 -3.11650
# 3 -87.726 41.844  c  1  3  0.33150
# 4 -87.719 41.849  d  3  4  0.68225
# 6 -87.722 41.838  e  2  1 -3.58675
# 7 -87.722 41.842  f  2  2 -3.00425

Note that ix, and iy could have also been found with a loop using findInterval, e.g. here\'s one example for the second row

findInterval(df$x[2], grid$x)
# 4
findInterval(df$y[2], grid$y)
# 2

Which matches ix and iy in df[2]

Footnote: (1) The fourth argument of vlookup was previously called \"match\", but after they introduced the ribbon it was renamed to \"[range_lookup]\".

回答7:

You could use mapvalues() from the plyr package.

Initial data:

dat <- data.frame(HouseType = c(\"Semi\", \"Single\", \"Row\", \"Single\", \"Apartment\", \"Apartment\", \"Row\"))

> dat
  HouseType
1      Semi
2    Single
3       Row
4    Single
5 Apartment
6 Apartment
7       Row

Lookup / crosswalk table:

lookup <- data.frame(type_text = c(\"Semi\", \"Single\", \"Row\", \"Apartment\"), type_num = c(1, 2, 3, 4))
> lookup
  type_text type_num
1      Semi        1
2    Single        2
3       Row        3
4 Apartment        4

Create the new variable:

dat$house_type_num <- plyr::mapvalues(dat$HouseType, from = lookup$type_text, to = lookup$type_num)

Or for simple replacements you can skip creating a long lookup table and do this directly in one step:

dat$house_type_num <- plyr::mapvalues(dat$HouseType,
                                      from = c(\"Semi\", \"Single\", \"Row\", \"Apartment\"),
                                      to = c(1, 2, 3, 4))

Result:

> dat
  HouseType house_type_num
1      Semi              1
2    Single              2
3       Row              3
4    Single              2
5 Apartment              4
6 Apartment              4
7       Row              3

回答8:

Using merge is different from lookup in Excel as it has potential to duplicate (multiply) your data if primary key constraint is not enforced in lookup table or reduce the number of records if you are not using all.x = T.

To make sure you don\'t get into trouble with that and lookup safely, I suggest two strategies.

First one is to make a check on a number of duplicated rows in lookup key:

safeLookup <- function(data, lookup, by, select = setdiff(colnames(lookup), by)) {
  # Merges data to lookup making sure that the number of rows does not change.
  stopifnot(sum(duplicated(lookup[, by])) == 0)
  res <- merge(data, lookup[, c(by, select)], by = by, all.x = T)
  return (res)
}

This will force you to de-dupe lookup dataset before using it:

baseSafe <- safeLookup(largetable, house.ids, by = \"HouseType\")
# Error: sum(duplicated(lookup[, by])) == 0 is not TRUE 

baseSafe<- safeLookup(largetable, unique(house.ids), by = \"HouseType\")
head(baseSafe)
# HouseType HouseTypeNo
# 1 Apartment           4
# 2 Apartment           4
# ...

Second option is to reproduce Excel behaviour by taking the first matching value from the lookup dataset:

firstLookup <- function(data, lookup, by, select = setdiff(colnames(lookup), by)) {
  # Merges data to lookup using first row per unique combination in by.
  unique.lookup <- lookup[!duplicated(lookup[, by]), ]
  res <- merge(data, unique.lookup[, c(by, select)], by = by, all.x = T)
  return (res)
}

baseFirst <- firstLookup(largetable, house.ids, by = \"HouseType\")

These functions are slightly different from lookup as they add multiple columns.