You could convert the underlying datetime64[ns] values to datetime64[s] values using astype: In [11]: df[‘Time’] = df[‘Time’].astype(‘datetime64[s]’) In [12]: df Out[12]: Record_ID Time 0 94704 2014-03-10 07:19:19 1 94705 2014-03-10 07:21:44 2 94706 2014-03-10 07:21:45 3 94707 2014-03-10 07:21:54 4 94708 2014-03-10 07:21:55 Note that since Pandas Series and DataFrames store all datetime values as … Read more
Creating dataframe from dictionary object. import pandas as pd data = [{‘name’: ‘vikash’, ‘age’: 27}, {‘name’: ‘Satyam’, ‘age’: 14}] df = pd.DataFrame.from_dict(data, orient=”columns”) df Out[4]: age name 0 27 vikash 1 14 Satyam If you have nested columns then you first need to normalize the data: data = [ { ‘name’: { ‘first’: ‘vikash’, ‘last’: … Read more
You can convert to string and extract the integer using regular expressions. df[‘B’].str.extract(‘(\d+)’).astype(int)
You can use clip. Apply to all columns of the data frame: df.clip(upper=15) Otherwise apply to selected columns as seen here: df.clip(upper=pd.Series({‘a’: 15}), axis=1)
there is even a shorter one 🙂 print df.groupby(‘name’).describe().unstack(1) Nothing beats one-liner: In [145]: print df.groupby(‘name’).describe().reset_index().pivot(index=’name’, values=”score”, columns=”level_1″)
You can use function map and select by function np.where more info print df # a b #0 aaa rrrr #1 bb k #2 ccc e #condition if condition is True then len column a else column b df[‘c’] = np.where(df[‘a’].map(len) > df[‘b’].map(len), df[‘a’].map(len), df[‘b’].map(len)) print df # a b c #0 aaa rrrr 4 … Read more
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