What is the equivalent to R's match() for python Pandas/numpy?

Question

I'm an R user and I cannot figure out the pandas equivalent of match(). I need use this function to iterate over a bunch of files, grab a key piece of info, and merge it back into the current data structure on 'url'. In R I'd do something like this:

logActions <- read.csv("data/logactions.csv")
logActions$class <- NA

files = dir("data/textContentClassified/")
for( i in 1:length(files)){
    tmp <- read.csv(files[i])
    logActions$class[match(logActions$url, tmp$url)] <- 
            tmp$class[match(tmp$url, logActions$url)]
}

I don't think I can use merge() or join(), as each will overwrite logActions$class each time. I can't use update() or combine_first() either, as neither have the necessary indexing capabilities. I also tried making a match() function based on this SO post, but cannot figure out how to get it to work with DataFrame objects. Apologies if I'm missing something obvious.

Here's some python code that summarizes my ineffectual attempts to do something like match() in pandas:

from pandas import *
left = DataFrame({'url': ['foo.com', 'foo.com', 'bar.com'], 'action': [0, 1, 0]})
left["class"] = NaN
right1 = DataFrame({'url': ['foo.com'], 'class': [0]})
right2 = DataFrame({'url': ['bar.com'], 'class': [ 1]})

# Doesn't work:
left.join(right1, on='url')
merge(left, right, on='url')

# Also doesn't work the way I need it to:
left = left.combine_first(right1)
left = left.combine_first(right2)
left 

# Also does something funky and doesn't really work the way match() does:
left = left.set_index('url', drop=False)
right1 = right1.set_index('url', drop=False)
right2 = right2.set_index('url', drop=False)

left = left.combine_first(right1)
left = left.combine_first(right2)
left

The desired output is:

    url  action  class
0   foo.com  0   0
1   foo.com  1   0
2   bar.com  0   1

BUT, I need to be able to call this over and over again so I can iterate over each file.

Answer 1

Note the existance of pandas.match which does precisely what R's match does.

Answer 2

Edit:

If url in all right dataframes re unique, you can make the right dataframe as a Series of class indexed by url, then you can get the class of every url in left by index it.

from pandas import *
left = DataFrame({'url': ['foo.com', 'bar.com', 'foo.com', 'tmp', 'foo.com'], 'action': [0, 1, 0, 2, 4]})
left["klass"] = NaN
right1 = DataFrame({'url': ['foo.com', 'tmp'], 'klass': [10, 20]})
right2 = DataFrame({'url': ['bar.com'], 'klass': [30]})

left["klass"] = left.klass.combine_first(right1.set_index('url').klass[left.url].reset_index(drop=True))
left["klass"] = left.klass.combine_first(right2.set_index('url').klass[left.url].reset_index(drop=True))

print left

Is this what you want?

import pandas as pd
left = pd.DataFrame({'url': ['foo.com', 'foo.com', 'bar.com'], 'action': [0, 1, 0]})
left["class"] = NaN
right1 = pd.DataFrame({'url': ['foo.com'], 'class': [0]})
right2 = pd.DataFrame({'url': ['bar.com'], 'class': [ 1]})

pd.merge(left.drop("class", axis=1), pd.concat([right1, right2]), on="url")

output:

   action      url  class
0       0  foo.com      0
1       1  foo.com      0
2       0  bar.com      1

if the class column in left is not all NaN, you can combine_fist it with the result.