Fix

Author: dsaxton

doc/source/user_guide/computation.rst

@@ -787,7 +787,11 @@ can even be omitted:
 
 .. ipython:: python
 
-   covs = df[["B", "C", "D"]].rolling(window=50).cov(df[["A", "B", "C"]], pairwise=True)
+   covs = (
+       df[["B", "C", "D"]]
+       .rolling(window=50)
+       .cov(df[["A", "B", "C"]], pairwise=True)
+   )
    covs.loc["2002-09-22":]
 
 .. ipython:: python

doc/source/user_guide/groupby.rst

@@ -267,7 +267,9 @@ the length of the ``groups`` dict, so it is largely just a convenience:
    height = np.random.normal(60, 10, size=n)
    time = pd.date_range("1/1/2000", periods=n)
    gender = np.random.choice(["male", "female"], size=n)
-   df = pd.DataFrame({"height": height, "weight": weight, "gender": gender}, index=time)
+   df = pd.DataFrame(
+       {"height": height, "weight": weight, "gender": gender}, index=time
+   )
 
 .. ipython:: python
 
@@ -767,7 +769,10 @@ For example, suppose we wished to standardize the data within each group:
    ts.head()
    ts.tail()
 
-   transformed = ts.groupby(lambda x: x.year).transform(lambda x: (x - x.mean()) / x.std())
+   transformed = ts.groupby(lambda x: x.year).transform(
+       lambda x: (x - x.mean()) / x.std()
+   )
+
 
 We would expect the result to now have mean 0 and standard deviation 1 within
 each group, which we can easily check:

doc/source/user_guide/missing_data.rst

@@ -400,7 +400,10 @@ You can also interpolate with a DataFrame:
 .. ipython:: python
 
    df = pd.DataFrame(
-       {"A": [1, 2.1, np.nan, 4.7, 5.6, 6.8], "B": [0.25, np.nan, np.nan, 4, 12.2, 14.4]}
+       {
+           "A": [1, 2.1, np.nan, 4.7, 5.6, 6.8],
+           "B": [0.25, np.nan, np.nan, 4, 12.2, 14.4],
+       }
    )
    df
    df.interpolate()

doc/source/user_guide/timeseries.rst

@@ -317,7 +317,9 @@ which can be specified. These are computed from the starting point specified by
 
 .. ipython:: python
 
-   pd.to_datetime([1349720105, 1349806505, 1349892905, 1349979305, 1350065705], unit="s")
+   pd.to_datetime(
+       [1349720105, 1349806505, 1349892905, 1349979305, 1350065705], unit="s"
+   )
 
    pd.to_datetime(
        [1349720105100, 1349720105200, 1349720105300, 1349720105400, 1349720105500],
@@ -707,7 +709,9 @@ If the timestamp string is treated as a slice, it can be used to index ``DataFra
 .. ipython:: python
     :okwarning:
 
-    dft_minute = pd.DataFrame({"a": [1, 2, 3], "b": [4, 5, 6]}, index=series_minute.index)
+    dft_minute = pd.DataFrame(
+        {"a": [1, 2, 3], "b": [4, 5, 6]}, index=series_minute.index
+    )
     dft_minute["2011-12-31 23"]
 
 
@@ -748,10 +752,11 @@ With no defaults.
 .. ipython:: python
 
    dft[
-       datetime.datetime(2013, 1, 1, 10, 12, 0): datetime.datetime(2013, 2, 28, 10, 12, 0)
+       datetime.datetime(2013, 1, 1, 10, 12, 0) : datetime.datetime(
+           2013, 2, 28, 10, 12, 0
+       )
    ]
 
-
 Truncating & fancy indexing
 ~~~~~~~~~~~~~~~~~~~~~~~~~~~
 
@@ -1036,8 +1041,15 @@ As an interesting example, let's look at Egypt where a Friday-Saturday weekend i
     # They also observe International Workers' Day so let's
     # add that for a couple of years
 
-    holidays = ["2012-05-01", datetime.datetime(2013, 5, 1), np.datetime64("2014-05-01")]
-    bday_egypt = pd.offsets.CustomBusinessDay(holidays=holidays, weekmask=weekmask_egypt)
+    holidays = [
+        "2012-05-01",
+        datetime.datetime(2013, 5, 1),
+        np.datetime64("2014-05-01"),
+    ]
+    bday_egypt = pd.offsets.CustomBusinessDay(
+        holidays=holidays,
+        weekmask=weekmask_egypt,
+    )
     dt = datetime.datetime(2013, 4, 30)
     dt + 2 * bday_egypt
 
@@ -1417,7 +1429,12 @@ An example of how holidays and holiday calendars are defined:
         rules = [
             USMemorialDay,
             Holiday("July 4th", month=7, day=4, observance=nearest_workday),
-            Holiday("Columbus Day", month=10, day=1, offset=pd.DateOffset(weekday=MO(2))),
+            Holiday(
+                "Columbus Day",
+                month=10,
+                day=1,
+                offset=pd.DateOffset(weekday=MO(2)),
+            ),
         ]
 
 
@@ -2279,15 +2296,25 @@ To return ``dateutil`` time zone objects, append ``dateutil/`` before the string
    rng_dateutil.tz
 
    # dateutil - utc special case
-   rng_utc = pd.date_range("3/6/2012 00:00", periods=3, freq="D", tz=dateutil.tz.tzutc())
+   rng_utc = pd.date_range(
+       "3/6/2012 00:00",
+       periods=3,
+       freq="D",
+       tz=dateutil.tz.tzutc(),
+   )
    rng_utc.tz
 
 .. versionadded:: 0.25.0
 
 .. ipython:: python
 
    # datetime.timezone
-   rng_utc = pd.date_range("3/6/2012 00:00", periods=3, freq="D", tz=datetime.timezone.utc)
+   rng_utc = pd.date_range(
+       "3/6/2012 00:00",
+       periods=3,
+       freq="D",
+       tz=datetime.timezone.utc,
+   )
    rng_utc.tz
 
 Note that the ``UTC`` time zone is a special case in ``dateutil`` and should be constructed explicitly
@@ -2440,10 +2467,18 @@ control over how they are handled.
 .. ipython:: python
 
    pd.Timestamp(
-       datetime.datetime(2019, 10, 27, 1, 30, 0, 0), tz="dateutil/Europe/London", fold=0
+       datetime.datetime(2019, 10, 27, 1, 30, 0, 0),
+       tz="dateutil/Europe/London",
+       fold=0,
    )
    pd.Timestamp(
-       year=2019, month=10, day=27, hour=1, minute=30, tz="dateutil/Europe/London", fold=1
+       year=2019,
+       month=10,
+       day=27,
+       hour=1,
+       minute=30,
+       tz="dateutil/Europe/London",
+       fold=1,
    )
 
 .. _timeseries.timezone_ambiguous:

doc/source/user_guide/visualization.rst

@@ -1453,7 +1453,11 @@ Here is an example of one way to easily plot group means with standard deviation
    )
 
    df3 = pd.DataFrame(
-       {"data1": [3, 2, 4, 3, 2, 4, 3, 2], "data2": [6, 5, 7, 5, 4, 5, 6, 5]}, index=ix3
+       {
+           "data1": [3, 2, 4, 3, 2, 4, 3, 2],
+           "data2": [6, 5, 7, 5, 4, 5, 6, 5],
+       },
+       index=ix3,
    )
 
    # Group by index labels and take the means and standard deviations