quanteda中dfm_groups无分组作用及dfm_weight(prop)频率含义问询

Let’s break down your two questions clearly:

1. Is dfm_group(groups=...) with no grouping variable equivalent to removing that line? #

When you call plot_topterms without passing a grouping variable, dfm_group(groups=...) will default to grouping all documents into a single group named all.

In terms of frequency calculation results, this is almost identical to removing the dfm_group line entirely—both approaches treat the entire corpus as one unified set when calculating proportional weights. The only minor difference is structural:

• If you remove dfm_group, the output from textstat_frequency won’t have a group column.
• If you keep dfm_group without a grouping variable, you’ll get a group column with every row set to all, which is why your facet_wrap renders a single panel labeled "all".

For your use case (plotting top terms across a single "all" group), the end visual and numerical results will be the same either way.

2. Does a relative frequency of 60 for "fun" mean 60% of documents contain the word? #

No, that’s a common misconception! Let’s clarify what dfm_weight(scheme='prop') actually calculates:

• scheme='prop' computes the term’s relative frequency within the entire corpus (or group)—meaning it’s the number of times the term appears divided by the total number of term occurrences in that corpus/group, multiplied by 100 (which is why you see 60 instead of 0.6).
• This is very different from document inclusion rate (the percentage of documents that contain the term at least once). To get that metric, you’d need to use dfm_weight(scheme='docfreq') instead, which counts how many documents each term appears in, then normalize by total documents.

For example:

• If your corpus has 100 total term occurrences, and "fun" appears 60 times, its proportional weight will be 60.
• But if only 30 out of 100 documents contain "fun" (even if it’s repeated multiple times in those 30), the document inclusion rate would be 30%, not 60%.

内容的提问来源于stack exchange，提问作者Ted Mosby