Fix intword for negative numbers

python-humanize · Jul 8, 2022 · bda2c5b · bda2c5b
1 parent 5dc4b82
commit bda2c5b
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Showing 2 changed files with 14 additions and 1 deletion.
diff --git a/src/humanize/number.py b/src/humanize/number.py
@@ -205,6 +205,11 @@ def intword(value: NumberOrString, format: str = "%.1f") -> str:
     except (TypeError, ValueError):
         return str(value)
 
+    is_negative = value < 0
+
+    if is_negative:
+        value *= -1
+
     if value < powers[0]:
         return str(value)
     for ordinal, power in enumerate(powers[1:], 1):
@@ -221,7 +226,8 @@ def intword(value: NumberOrString, format: str = "%.1f") -> str:
                 return (
                     " ".join([format, _ngettext(singular, plural, math.ceil(chopped))])
                 ) % chopped
-    return str(value)
+
+    return "-" + str(value) if is_negative else str(value)
 
 
 def apnumber(value: NumberOrString) -> str:

diff --git a/tests/test_number.py b/tests/test_number.py
@@ -75,24 +75,31 @@ def test_intword_powers() -> None:
 @pytest.mark.parametrize(
     "test_args, expected",
     [
+        (["0"], "0"),
         (["100"], "100"),
+        (["-100"], "-100"),
         (["1000"], "1.0 thousand"),
         (["12400"], "12.4 thousand"),
         (["12490"], "12.5 thousand"),
         (["1000000"], "1.0 million"),
+        (["-1000000"], "-1.0 million"),
         (["1200000"], "1.2 million"),
         (["1290000"], "1.3 million"),
         (["999999999"], "1.0 billion"),
         (["1000000000"], "1.0 billion"),
+        (["-1000000000"], "-1.0 billion"),
         (["2000000000"], "2.0 billion"),
         (["999999999999"], "1.0 trillion"),
         (["1000000000000"], "1.0 trillion"),
         (["6000000000000"], "6.0 trillion"),
+        (["-6000000000000"], "-6.0 trillion"),
         (["999999999999999"], "1.0 quadrillion"),
         (["1000000000000000"], "1.0 quadrillion"),
         (["1300000000000000"], "1.3 quadrillion"),
+        (["-1300000000000000"], "-1.3 quadrillion"),
         (["3500000000000000000000"], "3.5 sextillion"),
         (["8100000000000000000000000000000000"], "8.1 decillion"),
+        (["-8100000000000000000000000000000000"], "-8.1 decillion"),
         ([None], "None"),
         (["1230000", "%0.2f"], "1.23 million"),
         ([10**101], "1" + "0" * 101),