I would love to fill na in pandas dataframe where two columns in the dataframe both has on the same row.
A B C
2 3 5
Nan nan 7
4 7 9
Nan 4 9
12 5 8
Nan Nan 6
In the above dataframe, I would love to replace just row where both column A and B has Nan with "Not Available"
Thus:
A B C
2 3 5
Not available not available 7
4 7 9
Nan 4 9
12 5 8
Not available not available 6
I have tried multiple approach but I am getting undesirable result
If want test only A and B columns use DataFrame.loc with mask test it missing value with DataFrame.all for test if both are match:
m = df[['A','B']].isna().all(axis=1)
df.loc[m, ['A','B']] = 'Not Available'
If need test any 2 columns first count number of missing values and is 2 use fillna:
m = df.isna().sum(axis=1).eq(1)
df = df.where(m, df.fillna('Not Available'))
print (df)
A B C
0 2 3 5
1 Not Available Not Available 7
2 4 7 9
3 NaN 4 9
4 12 5 8
5 Not Available Not Available 6
Related
I have a Dataframe as shown below
A B C D
0 1 2 3.3 4
1 NaT NaN NaN NaN
2 NaT NaN NaN NaN
3 5 6 7 8
4 NaT NaN NaN NaN
5 NaT NaN NaN NaN
6 9 1 2 3
7 NaT NaN NaN NaN
8 NaT NaN NaN NaN
I need to copy the first row values (1,2,3,4) till the non-null row with index 2. Then, copy row values (5,6,7,8) till the non-null row with index 5 and copy (9,1,2,3) till row with index 8 and so on. Is there any way to do this in Python or Pandas. Quick help appreciated! Also is necessary not replace column D
Column C ffill gives 3.3456 as value for next row
Expected Output:
A B C D
0 1 2 3.3 4
1 1 2 3.3 NaN
2 1 2 3.3 NaN
3 5 6 7 8
4 5 6 7 NaN
5 5 6 7 NaN
6 9 1 2 3
7 9 1 2 NaN
8 9 1 2 NaN
Question was changed, so for forward filling all columns without D use Index.difference with ffill for columns names in list:
cols = df.columns.difference(['D'])
df[cols] = df[cols].ffill()
Or create mask for all columns names without D:
mask = df.columns != 'D'
df.loc[:, mask] = df.loc[:, mask].ffill()
EDIT: I cannot replicate your problem:
df = pd.DataFrame({'a':[2114.201789, np.nan, np.nan, 1]})
print (df)
a
0 2114.201789
1 NaN
2 NaN
3 1.000000
print (df.ffill())
a
0 2114.201789
1 2114.201789
2 2114.201789
3 1.000000
Having two data frames:
df1 = pd.DataFrame({'a':[1,2,3],'b':[4,5,6]})
a b
0 1 4
1 2 5
2 3 6
df2 = pd.DataFrame({'c':[7],'d':[8]})
c d
0 7 8
The goal is to add all df2 column values to df1, repeated and create the following result. It is assumed that both data frames do not share any column names.
a b c d
0 1 4 7 8
1 2 5 7 8
2 3 6 7 8
If there are strings columns names is possible use DataFrame.assign with unpack Series created by selecing first row of df2:
df = df1.assign(**df2.iloc[0])
print (df)
a b c d
0 1 4 7 8
1 2 5 7 8
2 3 6 7 8
Another idea is repeat values by df1.index with DataFrame.reindex and use DataFrame.join (here first index value of df2 is same like first index value of df1.index):
df = df1.join(df2.reindex(df1.index, method='ffill'))
print (df)
a b c d
0 1 4 7 8
1 2 5 7 8
2 3 6 7 8
If no missing values in original df is possible use forward filling missing values in last step, but also are types changed to floats, thanks #Dishin H Goyan:
df = df1.join(df2).ffill()
print (df)
a b c d
0 1 4 7.0 8.0
1 2 5 7.0 8.0
2 3 6 7.0 8.0
I have a data frame where there are several groups of numeric series where the values are cumulative. Consider the following:
df = pd.DataFrame({'Cat': ['A', 'A','A','A', 'B','B','B','B'], 'Indicator': [1,2,3,4,1,2,3,4], 'Cumulative1': [1,3,6,7,2,4,6,9], 'Cumulative2': [1,3,4,6,1,5,7,12]})
In [74]:df
Out[74]:
Cat Cumulative1 Cumulative2 Indicator
0 A 1 1 1
1 A 3 3 2
2 A 6 4 3
3 A 7 6 4
4 B 2 1 1
5 B 4 5 2
6 B 6 7 3
7 B 9 12 4
I need to create discrete series for Cumulative1 and Cumulative2, with starting point being the earliest entry in 'Indicator'.
my Approach is to use diff()
In[82]: df['Discrete1'] = df.groupby('Cat')['Cumulative1'].diff()
Out[82]: df
Cat Cumulative1 Cumulative2 Indicator Discrete1
0 A 1 1 1 NaN
1 A 3 3 2 2.0
2 A 6 4 3 3.0
3 A 7 6 4 1.0
4 B 2 1 1 NaN
5 B 4 5 2 2.0
6 B 6 7 3 2.0
7 B 9 12 4 3.0
I have 3 questions:
How do I avoid the NaN in an elegant/Pythonic way? The correct values are to be found in the original Cumulative series.
Secondly, how do I elegantly apply this computation to all series, say -
cols = ['Cumulative1', 'Cumulative2']
Thirdly, I have a lot of data that needs this computation -- is this the most efficient way?
You do not want to avoid NaNs, you want to fill them with the start values from the "cumulative" column:
df['Discrete1'] = df['Discrete1'].combine_first(df['Cumulative1'])
To apply the operation to all (or select) columns, broadcast it to all columns of interest:
sources = 'Cumulative1', 'Cumulative2'
targets = ["Discrete" + x[len('Cumulative'):] for x in sources]
df[targets] = df.groupby('Cat')[sources].diff()
You still have to condition the NaNs in a loop:
for s,t in zip(sources, targets):
df[t] = df[t].combine_first(df[s])
I need to remove all rows from a pandas.DataFrame, which satisfy an unusual condition.
In case there is an exactly the same row, except for it has Nan value in column "C", I want to remove this row.
Given a table:
A B C D
1 2 NaN 3
1 2 50 3
10 20 NaN 30
5 6 7 8
I need to remove the first row, since it has Nan in column C, but there is absolutely same row (second) with real value in column C.
However, 3rd row must stay, because there're no rows with same A, B and D values as it has.
How do you perform this using pandas? Thank you!
You can achieve in using drop_duplicates.
Initial DataFrame:
df=pd.DataFrame(columns=['a','b','c','d'], data=[[1,2,None,3],[1,2,50,3],[10,20,None,30],[5,6,7,8]])
df
a b c d
0 1 2 NaN 3
1 1 2 50 3
2 10 20 NaN 30
3 5 6 7 8
Then you can sort DataFrame by column C. This will drop NaNs to the bottom of column:
df = df.sort_values(['c'])
df
a b c d
3 5 6 7 8
1 1 2 50 3
0 1 2 NaN 3
2 10 20 NaN 30
And then remove duplicates selecting taken into account columns ignoring C and keeping first catched row:
df1 = df.drop_duplicates(['a','b','d'], keep='first')
a b c d
3 5 6 7 8
1 1 2 50 3
2 10 20 NaN 30
But it will be valid only if NaNs are in column C.
You can try fillna along with drop_duplicates
df.bfill().ffill().drop_duplicates(subset=['A', 'B', 'D'], keep = 'last')
This will handle the scenario such as A, B and D values are same but C has non-NaN values in both the rows.
You get
A B C D
1 1 2 50 3
2 10 20 Nan 30
3 5 6 7 8
This feels right to me
notdups = ~df.duplicated(df.columns.difference(['C']), keep=False)
notnans = df.C.notnull()
df[notdups | notnans]
A B C D
1 1 2 50.0 3
2 10 20 NaN 30
3 5 6 7.0 8
Considering the following dataframe:
index group signal
1 1 1
2 1 NAN
3 1 NAN
4 1 -1
5 1 NAN
6 2 NAN
7 2 -1
8 2 NAN
9 3 NAN
10 3 NAN
11 3 NAN
12 4 1
13 4 NAN
14 4 NAN
I want to modify the signals by ffill NANs in each group so that I can have the following dataframe:
index group signal
1 1 1
2 1 1
3 1 1
4 1 -1
5 1 -1
6 2 NAN
7 2 -1
8 2 -1
9 3 NAN
10 3 NAN
11 3 NAN
12 4 1
13 4 1
14 4 1
The dataframe is big (around 800,000 rows with about 16,000 different groups) and currently I put it into a groupby object and try to modify each group there, which is very slow. Then I tried to convert it into a pivot_table and ffill() there, but the dataframe is simple too large and the program gives errors. Any suggestions? Thank you!
Can you try out this
data_group = data.groupby('group').apply(lambda v: v.fillna(method='ffill'))
I think in your data NAN is a string. Its not a empty element. Empty data will appear as NaN. If it is a string, do a replacement of NAN. Like
data_group = data.groupby('group').apply(lambda v: v.replace('NAN', float('nan')).fillna(method='ffill'))
Or a better version as Jeff suggested
data['signal'] = data['signal'].replace('NAN', float('nan'))
data = data.groupby('group').ffill()