To delete multiple columns at the same time in pandas, you could specify the column names as shown below. The option inplace=True is needed if one wants the change affected column in the same dataframe. Otherwise remove it. flight_data_copy.drop([‘TailNum’, ‘OriginStateFips’, ‘DestStateFips’, ‘Diverted’], axis=1, inplace=True) Source: Python Pandas – Deleting multiple series from a data frame … Read more
Here are few ways to add an empty column to an empty dataframe: df=pd.DataFrame(columns=[‘a’]) df[‘b’] = None df = df.assign(c=None) df = df.assign(d=df[‘a’]) df[‘e’] = pd.Series(index=df.index) df = pd.concat([df,pd.DataFrame(columns=list(‘f’))]) print(df) Output: Empty DataFrame Columns: [a, b, c, d, e, f] Index: [] I hope it helps.
reset_index by default does not modify the DataFrame; it returns a new DataFrame with the reset index. If you want to modify the original, use the inplace argument: df.reset_index(drop=True, inplace=True). Alternatively, assign the result of reset_index by doing df = df.reset_index(drop=True).
Spark 2.0 This is issue has been resolved in Spark 2.0 with SPARK-14241. Another similar issue has been resolved in Spark 2.1 with SPARK-14393 Spark 1.x Problem you experience is rather subtle but can be reduced to a simple fact monotonically_increasing_id is an extremely ugly function. It is clearly not pure and its value depends … Read more
If I understand correctly you can add query(“Points > 100”): df = pd.DataFrame({‘Points’:[50,20,38,90,0, np.Inf], ‘Player’:[‘a’,’a’,’a’,’s’,’s’,’s’]}) print (df) Player Points 0 a 50.000000 1 a 20.000000 2 a 38.000000 3 s 90.000000 4 s 0.000000 5 s inf points_series = df.query(“Points < inf”).groupby(“Player”).agg({“Points”: “sum”})[‘Points’] print (points_series) a = points_series[points_series > 100] print (a) Player a 108.0 … Read more
For improve performance is possible use numexpr: import numexpr np.random.seed(125) N = 40000000 df = pd.DataFrame({‘A’:np.random.randint(10, size=N)}) def ne(df): x = df.A.values return df[numexpr.evaluate(‘(x > 5)’)] print (ne(df)) In [138]: %timeit (ne(df)) 1 loop, best of 3: 494 ms per loop In [139]: %timeit df[df.A > 5] 1 loop, best of 3: 536 ms per … Read more
It might be easiest to turn your Series into a DataFrame and use Pandas’ groupby functionality (if you already have a DataFrame then skip straight to adding another column below). If your Series is called s, then turn it into a DataFrame like so: >>> df = pd.DataFrame({‘Timestamp’: s.index, ‘Category’: s.values}) >>> df Category Timestamp … Read more
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