dfm | dfm works best when it 's included in your dotfiles | Configuration Management library

You don't need to create your dictionary from the corpus - instead, create a single dictionary entry for your locality list, and look that up to generate a count of each locality. You can then count them by compiling the dfm, and then converting the feature of that dictionary key into a logical to get the vector you want.

This is a case where knowing the value of return objects in R is the key to obtaining the result you want. Specifically, you need to know what stopwords() returns, as well as what it is expected as its first argument.

stopwords(language = "sp") returns a character vector of Spanish stopwords, using the default source = "snowball" list. (See ?stopwords for full details.)

So if you want to remove the default Spanish list plus your own words, you concatenate the returned character vector with additional elements. This is what you have done in creating all_stops.

So to remove all_stops -- and here, using the quanteda v3 suggested usage -- you simply do the following:

First thing, change ct into pd.datetime. I am not certain about the compatibility between python datetime and pandas datetime. You can basically filter after that.

For instance, code I use for changing now into pd.datetime is:

It looks like some of your documents are empty, in the sense that they contain no counts of any feature.

You can remove them with:

Both derivatives got some signs mixed up:

We changed the usage of the groups argument in v3 to make it more standard.

From news(Version >= "3.0", package = "quanteda"):

We have added non-standard evaluation for by and groups arguments to access object docvars:

• The *_sample() functions' argument by, and groups in the *_group() functions, now take unquoted document variable (docvar) names directly, similar to the way the subset argument works in the *_subset() functions.
• Quoted docvar names no longer work, as these will be evaluated literally.
• The by = "document" formerly sampled from docid(x), but this functionality is now removed. Instead, use by = docid(x) to replicate this functionality.
• For groups, the default is now docid(x), which is now documented more completely. See ?groups and ?docid.

So, to get the previous behaviour, you would want to use:

You need to use a cast like this:

Splitting the strings at the underscore and sort them alphabetically, then use this list to identify duplicates and apply it to the original list:

I can reproduce your problem. I found a workaround: use the display rectangle of the selected item: