@@ -464,7 +464,10 @@ which we illustrate:
464464 {" A" : [1.0 , np.nan, 3.0 , 5.0 , np.nan], " B" : [np.nan, 2.0 , 3.0 , np.nan, 6.0 ]}
465465 )
466466 df2 = pd.DataFrame(
467- {" A" : [5.0 , 2.0 , 4.0 , np.nan, 3.0 , 7.0 ], " B" : [np.nan, np.nan, 3.0 , 4.0 , 6.0 , 8.0 ]}
467+ {
468+ " A" : [5.0 , 2.0 , 4.0 , np.nan, 3.0 , 7.0 ],
469+ " B" : [np.nan, np.nan, 3.0 , 4.0 , 6.0 , 8.0 ],
470+ }
468471 )
469472 df1
470473 df2
@@ -712,7 +715,10 @@ Similarly, you can get the most frequently occurring value(s), i.e. the mode, of
712715 s5 = pd.Series([1 , 1 , 3 , 3 , 3 , 5 , 5 , 7 , 7 , 7 ])
713716 s5.mode()
714717 df5 = pd.DataFrame(
715- {" A" : np.random.randint(0 , 7 , size = 50 ), " B" : np.random.randint(- 10 , 15 , size = 50 )}
718+ {
719+ " A" : np.random.randint(0 , 7 , size = 50 ),
720+ " B" : np.random.randint(- 10 , 15 , size = 50 ),
721+ }
716722 )
717723 df5.mode()
718724
@@ -1192,7 +1198,9 @@ to :ref:`merging/joining functionality <merging>`:
11921198
11931199.. ipython :: python
11941200
1195- s = pd.Series([" six" , " seven" , " six" , " seven" , " six" ], index = [" a" , " b" , " c" , " d" , " e" ])
1201+ s = pd.Series(
1202+ [" six" , " seven" , " six" , " seven" , " six" ], index = [" a" , " b" , " c" , " d" , " e" ]
1203+ )
11961204 t = pd.Series({" six" : 6.0 , " seven" : 7.0 })
11971205 s
11981206 s.map(t)
@@ -1494,7 +1502,9 @@ labels).
14941502
14951503 df = pd.DataFrame(
14961504 {" x" : [1 , 2 , 3 , 4 , 5 , 6 ], " y" : [10 , 20 , 30 , 40 , 50 , 60 ]},
1497- index = pd.MultiIndex.from_product([[" a" , " b" , " c" ], [1 , 2 ]], names = [" let" , " num" ]),
1505+ index = pd.MultiIndex.from_product(
1506+ [[" a" , " b" , " c" ], [1 , 2 ]], names = [" let" , " num" ]
1507+ ),
14981508 )
14991509 df
15001510 df.rename_axis(index = {" let" : " abc" })
@@ -1803,7 +1813,9 @@ used to sort a pandas object by its index levels.
18031813 }
18041814 )
18051815
1806- unsorted_df = df.reindex(index = [" a" , " d" , " c" , " b" ], columns = [" three" , " two" , " one" ])
1816+ unsorted_df = df.reindex(
1817+ index = [" a" , " d" , " c" , " b" ], columns = [" three" , " two" , " one" ]
1818+ )
18071819 unsorted_df
18081820
18091821 # DataFrame
@@ -1849,7 +1861,9 @@ to use to determine the sorted order.
18491861
18501862.. ipython :: python
18511863
1852- df1 = pd.DataFrame({" one" : [2 , 1 , 1 , 1 ], " two" : [1 , 3 , 2 , 4 ], " three" : [5 , 4 , 3 , 2 ]})
1864+ df1 = pd.DataFrame(
1865+ {" one" : [2 , 1 , 1 , 1 ], " two" : [1 , 3 , 2 , 4 ], " three" : [5 , 4 , 3 , 2 ]}
1866+ )
18531867 df1.sort_values(by = " two" )
18541868
18551869 The ``by `` parameter can take a list of column names, e.g.:
@@ -1994,7 +2008,9 @@ all levels to ``by``.
19942008
19952009.. ipython :: python
19962010
1997- df1.columns = pd.MultiIndex.from_tuples([(" a" , " one" ), (" a" , " two" ), (" b" , " three" )])
2011+ df1.columns = pd.MultiIndex.from_tuples(
2012+ [(" a" , " one" ), (" a" , " two" ), (" b" , " three" )]
2013+ )
19982014 df1.sort_values(by = (" a" , " two" ))
19992015
20002016
@@ -2245,7 +2261,11 @@ to the correct type.
22452261 import datetime
22462262
22472263 df = pd.DataFrame(
2248- [[1 , 2 ], [" a" , " b" ], [datetime.datetime(2016 , 3 , 2 ), datetime.datetime(2016 , 3 , 2 )]]
2264+ [
2265+ [1 , 2 ],
2266+ [" a" , " b" ],
2267+ [datetime.datetime(2016 , 3 , 2 ), datetime.datetime(2016 , 3 , 2 )],
2268+ ]
22492269 )
22502270 df = df.T
22512271 df
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