How to apply grepl for data frame

Question

I want to use grepl for multiple patterns defined as a data frame. df_sen is presented as

sentence
"She would like to go there"
"I had it few days ago"
"We have spent few millions"

df_triggers is presented as follows:

trigger
few days
few millions

And I want to create a matrix where sentence x triggers and on the intersection to see 1 if trigger was found in a sentence and 0 if it was not.

I have tried to do it like this:

matrix <- grepl(df_triggers$trigger, df_sen$sentence)

But I see the error message that I have more than 1 pattern in grepl().

The desired output is:

                                 few days    few millions
"She would like to go there"        0              0
"I had it few days ago"             1              0
"We have spent few millions         0              1

And the thing is that I have 2500 sentences. So I cannot implement it like `few days | few millions` — ZverArt, May 12 '17 at 11:15
Try an sapply loop over the patterns: `sapply(df_triggers$trigger, grepl, df_sen$sentence)` — talat, May 12 '17 at 11:19
It would be better if you could post the solution as an answer below the question. Thanks! — djhurio, May 12 '17 at 12:07

score 2 · Accepted Answer · edited May 12 '17 at 12:17

2

sapply(df_triggers$trigger, grepl, df_sen$sentence)

from @docendodiscimus worked.

edited May 12 '17 at 12:17

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answered May 12 '17 at 12:14

ZverArt

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How to apply grepl for data frame

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