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Pandas: get rid off rows where there is no value


By : kuangkai
Date : October 18 2020, 06:10 AM
wish of those help did you try using the dropna method?
df.dropna(how='all') , the 'all' will drop if all the values are NA.
code :


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pandas apply function that returns multiple values to rows in pandas dataframe


By : Sharat Deb
Date : March 29 2020, 07:55 AM
like below fixes the issue I have a dataframe with a timeindex and 3 columns containing the coordinates of a 3D vector: , Just return a list instead of tuple.
code :
In [81]: df
Out[81]: 
                            x         y         z
ts                                               
2014-05-15 10:38:00  0.120117  0.987305  0.116211
2014-05-15 10:39:00  0.117188  0.984375  0.122070
2014-05-15 10:40:00  0.119141  0.987305  0.119141
2014-05-15 10:41:00  0.116211  0.984375  0.120117
2014-05-15 10:42:00  0.119141  0.983398  0.118164

[5 rows x 3 columns]

In [82]: def myfunc(args):
   ....:        e=args[0] + 2*args[1]
   ....:        f=args[1]*args[2] +1
   ....:        g=args[2] + args[0] * args[1]
   ....:        return [e,f,g]
   ....: 

In [83]: df.apply(myfunc ,axis=1)
Out[83]: 
                            x         y         z
ts                                               
2014-05-15 10:38:00  2.094727  1.114736  0.234803
2014-05-15 10:39:00  2.085938  1.120163  0.237427
2014-05-15 10:40:00  2.093751  1.117629  0.236770
2014-05-15 10:41:00  2.084961  1.118240  0.234512
2014-05-15 10:42:00  2.085937  1.116202  0.235327

Pandas: How to apply a passed conditional operator to select rows in pandas?


By : gerald hewes
Date : March 29 2020, 07:55 AM
I hope this helps . I think need use operator in (), what is same like operator.lt(column, value):
code :
# helper.py
def get_milage(operator, condition):
    return df[operator(df.Milage, condition)]

Python: how to drop rows in Pandas if two columns don't appear in another pandas column?


By : restalion
Date : March 29 2020, 07:55 AM
Hope this helps I have two dataframes df and df1. df contains name and attributes of people. , Use isin -
code :
df1[df1[['Name1', 'Name2']].isin(df.Name).all(1)]

#    Name1 Name2  c
#2   Frank  John  1
#3  Emilie  Jack  3
df1[df1.Name1.isin(df.Name) & df1.Name2.isin(df.Name)]

#    Name1 Name2  c
#2   Frank  John  1
#3  Emilie  Jack  3

pandas - Iterate dataframe rows, modify them, and rebuild a data frame in a for loop pandas


By : AMod
Date : March 29 2020, 07:55 AM
it fixes the issue Not sure if this is the best way, but it can be done by defining a function as below to create new 2 row for every row in old df:
code :
def get_new(row, awindow, bwindow):                                
    new_row_A = {}         
    new_row_A['CHROMOSOME'] = row['CHROMOSOME']                        
    new_row_A['START'] = row['START']-awindow
    new_row_A['END'] = row['START']
    new_row_A['LABEL'] = 'AVN_DEL_'+str(row.name)+'_A'
    new_row_B = {}
    new_row_B['CHROMOSOME'] = row['CHROMOSOME']
    new_row_B['START'] = row['END']
    new_row_B['END'] = row['END']+bwindow
    new_row_B['LABEL'] = 'AVN_DEL_'+str(row.name)+'_B'
    return [new_row,new_row_B]
awindow = 500
bwindow = 500
new_df = pd.DataFrame()
for new_row in df.apply(lambda row: get_new(row, awindow, bwindow), axis=1):
    new_df = new_df.append(pd.DataFrame(new_row))
new_df.reset_index(drop=True, inplace=True)

Python pandas: filtering rows based on time criteria using pandas


By : Ileaf Infopark Thapa
Date : March 29 2020, 07:55 AM
will help you Use pd.Grouper:
code :
n=5
df.groupby(pd.Grouper(key = 'Time', freq=f'{n} min')).first()

                      Amount   Price
Time                                 
2010-01-01 00:00:00    0.36  13924.98
2010-01-01 00:05:00    0.07  13907.59
2010-01-01 00:10:00    0.03  13909.99
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