when measuring BPC, divide by ln(2) to get into the right base

proger · Apr 26, 2023 · 1c5dc38 · 1c5dc38
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commit 1c5dc38
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diff --git a/examples/exp/ppl/BPC b/examples/exp/ppl/BPC
@@ -1,12 +1,12 @@
 data/flair-uk-forward.ppl.tsv sentences 28643
 data/flair-uk-forward.ppl.tsv nll_mean 139.6773665773955
-data/flair-uk-forward.ppl.tsv bpc 0.8214119255908473
+data/flair-uk-forward.ppl.tsv bpc 1.7096613025528762
 exp/ppl/small.tsv sentences 28643
 exp/ppl/small.tsv nll_mean 118.76219278015107
-exp/ppl/small.tsv bpc 0.7222614857594851
+exp/ppl/small.tsv bpc 1.503292652634816
 exp/ppl/medium.tsv sentences 28643
 exp/ppl/medium.tsv nll_mean 115.0707072714907
-exp/ppl/medium.tsv bpc 0.6998114303527089
+exp/ppl/medium.tsv bpc 1.4565658036892946
 exp/ppl/large.tsv sentences 28643
 exp/ppl/large.tsv nll_mean 113.01209246430402
-exp/ppl/large.tsv bpc 0.6872915942213289
+exp/ppl/large.tsv bpc 1.4305074051181672
diff --git a/examples/scripts/evaluate_nll.py b/examples/scripts/evaluate_nll.py
@@ -24,7 +24,7 @@
     df = df.loc[df.index.intersection(idf.index)]
 
 nll = np.log(df.ppl.to_numpy()) * df.sentence_len.to_numpy()
-nll2 = nll / np.log2(np.e)
+nll2 = nll / np.log(2)
 
 char_len = df.text.str.len().to_numpy()
 N = np.sum(char_len)