I have a dataframe as example:
A B C
0 1
1 1
2 1
3 1 2
4 1 2
5 1 2
6 2 3
7 2 3
8 2 3
9 3
10 3
11 3
And I would like to remove nan values of each column to get the result:
A B C
0 1 2 3
1 1 2 3
2 1 2 3
3 1 2 3
4 1 2 3
5 1 2 3
Do I have an easy way to do that?
You can apply a custom sorting function for each column that doesn't actually sort numerically, it justs moves all the NaN values to the end of the column. Then, dropna:
df = df.apply(lambda x: sorted(x, key=lambda v: isinstance(v, float) and np.isnan(v))).dropna()
Output:
>>> df
A B C
0 1.0 2.0 3.0
1 1.0 2.0 3.0
2 1.0 2.0 3.0
3 1.0 2.0 3.0
4 1.0 2.0 3.0
5 1.0 2.0 3.0
Given
>>> df
A B C
0 1.0 NaN NaN
1 1.0 NaN NaN
2 1.0 NaN NaN
3 1.0 2.0 NaN
4 1.0 2.0 NaN
5 1.0 2.0 NaN
6 NaN 2.0 3.0
7 NaN 2.0 3.0
8 NaN 2.0 3.0
9 NaN NaN 3.0
10 NaN NaN 3.0
11 NaN NaN 3.0
use
>>> df.apply(lambda s: s.dropna().to_numpy())
A B C
0 1.0 2.0 3.0
1 1.0 2.0 3.0
2 1.0 2.0 3.0
3 1.0 2.0 3.0
4 1.0 2.0 3.0
5 1.0 2.0 3.0
Related
The objective is to fill NaN with respect to two columns (i.e., a, b) .
a b c d
2,0,1,4
5,0,5,6
6,0,1,1
1,1,1,4
4,1,5,6
5,1,5,6
6,1,1,1
1,2,2,3
6,2,5,6
Such that, there should be continous value of between 1 to 6 for the column a for a fixed value in column b. Then, the other rows assigned to nan.
The code snippet does the trick
import numpy as np
import pandas as pd
maxval_col_a=6
lowval_col_a=1
maxval_col_b=2
lowval_col_b=0
r=list(range(lowval_col_b,maxval_col_b+1))
df=pd.DataFrame(np.column_stack([[2,5,6,1,4,5,6,1,6,],
[0,0,0,1,1,1,1,2,2,], [1,5,1,1,5,5,1,2,5,],[4,6,1,4,6,6,1,3,6,]]),columns=['a','b','c','d'])
all_df=[]
for idx in r:
k=df.loc[df['b']==idx].set_index('a').reindex(range(lowval_col_a, maxval_col_a+1, 1)).reset_index()
k['b']=idx
all_df.append(k)
df=pd.concat(all_df)
But, I am curious whether there are more efficient and better way of doing this with Pandas.
The expected output
a b c d
0 1 0 NaN NaN
1 2 0 1.0 4.0
2 3 0 NaN NaN
3 4 0 NaN NaN
4 5 0 5.0 6.0
5 6 0 1.0 1.0
0 1 1 1.0 4.0
1 2 1 NaN NaN
2 3 1 NaN NaN
3 4 1 5.0 6.0
4 5 1 5.0 6.0
5 6 1 1.0 1.0
0 1 2 2.0 3.0
1 2 2 NaN NaN
2 3 2 NaN NaN
3 4 2 NaN NaN
4 5 2 NaN NaN
5 6 2 5.0 6.0
Create the cartesian product of combinations:
mi = pd.MultiIndex.from_product([df['b'].unique(), range(1, 7)],
names=['b', 'a']).swaplevel()
out = df.set_index(['a', 'b']).reindex(mi).reset_index()
print(out)
# Output
a b c d
0 1 0 NaN NaN
1 2 0 1.0 4.0
2 3 0 NaN NaN
3 4 0 NaN NaN
4 5 0 5.0 6.0
5 6 0 1.0 1.0
6 1 1 1.0 4.0
7 2 1 NaN NaN
8 3 1 NaN NaN
9 4 1 5.0 6.0
10 5 1 5.0 6.0
11 6 1 1.0 1.0
12 1 2 2.0 3.0
13 2 2 NaN NaN
14 3 2 NaN NaN
15 4 2 NaN NaN
16 5 2 NaN NaN
17 6 2 5.0 6.0
First create a multindex with cols [a,b] then a new multindex with all the combinations and then you reindex with the new multindex:
(showing all steps)
# set both a and b as index (it's a multiindex)
df.set_index(['a','b'],drop=True,inplace=True)
# create the new multindex
new_idx_a=np.tile(np.arange(0,6+1),3)
new_idx_b=np.repeat([0,1,2],6+1)
new_multidx=pd.MultiIndex.from_arrays([new_idx_a,
new_idx_b])
# reindex
df=df.reindex(new_multidx)
# convert the multindex back to columns
df.index.names=['a','b']
df.reset_index()
results:
a b c d
0 0 0 NaN NaN
1 1 0 NaN NaN
2 2 0 1.0 4.0
3 3 0 NaN NaN
4 4 0 NaN NaN
5 5 0 5.0 6.0
6 6 0 1.0 1.0
7 0 1 NaN NaN
8 1 1 1.0 4.0
9 2 1 NaN NaN
10 3 1 NaN NaN
11 4 1 5.0 6.0
12 5 1 5.0 6.0
13 6 1 1.0 1.0
14 0 2 NaN NaN
15 1 2 2.0 3.0
16 2 2 NaN NaN
17 3 2 NaN NaN
18 4 2 NaN NaN
19 5 2 NaN NaN
20 6 2 5.0 6.0
We can do it by using a groupby on the column b, then set a as index and add the missing values of a using numpy.arange.
To finish, reset the index to get the expected result :
import numpy as np
df.groupby('b').apply(lambda x : x.set_index('a').reindex(np.arange(1, 7))).drop('b', 1).reset_index()
Output :
b a c d
0 0 1 NaN NaN
1 0 2 1.0 4.0
2 0 3 NaN NaN
3 0 4 NaN NaN
4 0 5 5.0 6.0
5 0 6 1.0 1.0
6 1 1 1.0 4.0
7 1 2 NaN NaN
8 1 3 NaN NaN
9 1 4 5.0 6.0
10 1 5 5.0 6.0
11 1 6 1.0 1.0
12 2 1 2.0 3.0
13 2 2 NaN NaN
14 2 3 NaN NaN
15 2 4 NaN NaN
16 2 5 NaN NaN
17 2 6 5.0 6.0
i'm in this situation,
my df is like that
A B
0 0.0 2.0
1 3.0 4.0
2 NaN 1.0
3 2.0 NaN
4 NaN 1.0
5 4.8 NaN
6 NaN 1.0
and i want to apply this line of code:
df['A'] = df['B'].fillna(df['A'])
and I expect a workflow and final output like that:
A B
0 2.0 2.0
1 4.0 4.0
2 1.0 1.0
3 NaN NaN
4 1.0 1.0
5 NaN NaN
6 1.0 1.0
A B
0 2.0 2.0
1 4.0 4.0
2 1.0 1.0
3 2.0 NaN
4 1.0 1.0
5 4.8 NaN
6 1.0 1.0
but I receive this error:
TypeError: Unsupported type Series
probably because each time there is an NA it tries to fill it with the whole series and not with the single element with the same index of the B column.
I receive the same error with a syntax like that:
df['C'] = df['B'].fillna(df['A'])
so the problem seems not to be the fact that I'm first changing the values of A with the ones of B and then trying to fill the "B" NA with the values of a column that is technically the same as B
I'm in a databricks environment and I'm working with koalas data frames but they work as the pandas ones.
can you help me?
Another option
Suppose the following dataset
import pandas as pd
import numpy as np
df = pd.DataFrame(data={'State':[1,2,3,4,5,6, 7, 8, 9, 10],
'Sno Center': ["Guntur", "Nellore", "Visakhapatnam", "Biswanath", "Doom-Dooma", "Guntur", "Labac-Silchar", "Numaligarh", "Sibsagar", "Munger-Jamalpu"],
'Mar-21': [121, 118.8, 131.6, 123.7, 127.8, 125.9, 114.2, 114.2, 117.7, 117.7],
'Apr-21': [121.1, 118.3, 131.5, np.NaN, 128.2, 128.2, 115.4, 115.1, np.NaN, 118.3]})
df
State Sno Center Mar-21 Apr-21
0 1 Guntur 121.0 121.1
1 2 Nellore 118.8 118.3
2 3 Visakhapatnam 131.6 131.5
3 4 Biswanath 123.7 NaN
4 5 Doom-Dooma 127.8 128.2
5 6 Guntur 125.9 128.2
6 7 Labac-Silchar 114.2 115.4
7 8 Numaligarh 114.2 115.1
8 9 Sibsagar 117.7 NaN
9 10 Munger-Jamalpu 117.7 118.3
Then
df.loc[(df["Mar-21"].notnull()) & (df["Apr-21"].isna()), "Apr-21"] = df["Mar-21"]
df
State Sno Center Mar-21 Apr-21
0 1 Guntur 121.0 121.1
1 2 Nellore 118.8 118.3
2 3 Visakhapatnam 131.6 131.5
3 4 Biswanath 123.7 123.7
4 5 Doom-Dooma 127.8 128.2
5 6 Guntur 125.9 128.2
6 7 Labac-Silchar 114.2 115.4
7 8 Numaligarh 114.2 115.1
8 9 Sibsagar 117.7 117.7
9 10 Munger-Jamalpu 117.7 118.3
IIUC:
try with max():
df['A']=df[['A','B']].max(axis=1)
output of df:
A B
0 2.0 2.0
1 4.0 4.0
2 1.0 1.0
3 2.0 NaN
4 1.0 1.0
5 4.8 NaN
6 1.0 1.0
I've got df as follows:
a b
0 1 NaN
1 2 NaN
2 1 1.0
3 4 NaN
4 9 1.0
5 6 NaN
6 5 2.0
7 8 NaN
8 9 2.0
I'd like fill nan's only between numbers to get df like this:
a b
0 1 NaN
1 2 NaN
2 1 1.0
3 4 1.0
4 9 1.0
5 6 NaN
6 5 2.0
7 8 2.0
8 9 2.0
and then create two new dataframes:
a b
2 1 1.0
3 4 1.0
4 9 1.0
a b
6 5 2.0
7 8 2.0
8 9 2.0
meaning select all columns and rows with fiiled out nan only.
My idea for first part, this with filling out nan is to create separate dataframe with row indexes like:
2 1.0
4 1.0
6 2.0
8 2.0
and based on that create range of row indexes to fill out.
My question is maybe there is, related to this part with replacing nan, more pythonic function to do this.
How about
df[df.b.ffill()==df.b.bfill()].ffill()
results in
# a b
# 2 1 1.0
# 3 4 1.0
# 4 9 1.0
# 6 5 2.0
# 7 8 2.0
# 8 9 2.0
Explanation:
df['c'] = df.b.ffill()
df['d'] = df.b.bfill()
# a b c d
# 0 1 NaN NaN 1.0
# 1 2 NaN NaN 1.0
# 2 1 1.0 1.0 1.0
# 3 4 NaN 1.0 1.0
# 4 9 1.0 1.0 1.0
# 5 6 NaN 1.0 2.0
# 6 5 2.0 2.0 2.0
# 7 8 NaN 2.0 2.0
# 8 9 2.0 2.0 2.0
I have a dataframe df
import pandas as pd
import numpy as np
df = pd.DataFrame({'A': [np.nan, 1, 2,np.nan,2,np.nan,np.nan],
'B': [10, np.nan, np.nan,5,np.nan,np.nan,7],
'C': [1,1,2,2,3,3,3]})
which looks like :
A B C
0 NaN 10.0 1
1 1.0 NaN 1
2 2.0 NaN 2
3 NaN 5.0 2
4 2.0 NaN 3
5 NaN NaN 3
6 NaN 7.0 3
I want to replace all the NAN values in column A and B with the value from other records which are from the same group as mentioned in column C.
My expected output is :
A B C
0 1.0 10.0 1
1 1.0 10.0 1
2 2.0 5.0 2
3 2.0 5.0 2
4 2.0 7.0 3
5 2.0 7.0 3
6 2.0 7.0 3
How can I do the same in pandas dataframe ?
Use GroupBy.apply with forward and back filling missing values:
df[['A','B']] = df.groupby('C')['A','B'].apply(lambda x: x.ffill().bfill())
print (df)
A B C
0 1.0 10.0 1
1 1.0 10.0 1
2 2.0 5.0 2
3 2.0 5.0 2
4 2.0 7.0 3
5 2.0 7.0 3
6 2.0 7.0 3
Suppose I produce the following using pandas:
df = pd.DataFrame(np.ones(25).reshape((5,5)),index = ['A','B','C','D','E'])
df1 = pd.DataFrame(np.ones(25).reshape((5,5))*2,index = ['A','B','C','D','E'])
df[2] = np.nan
df1[3] = np.nan
df[4] = np.nan
df1[4] = np.nan
df2 = df+df1
print(df2)
0 1 2 3 4
A 3.0 3.0 NaN NaN NaN
B 3.0 3.0 NaN NaN NaN
C 3.0 3.0 NaN NaN NaN
D 3.0 3.0 NaN NaN NaN
E 3.0 3.0 NaN NaN NaN
What do I have to do to get this instead?
0 1 2 3 4
A 3 3 2 1 NaN
B 3 3 2 1 NaN
C 3 3 2 1 NaN
D 3 3 2 1 NaN
E 3 3 2 1 NaN
Use the fill_value argument of the DataFrame.add method:
fill_value : None or float value, default None Fill missing (NaN)
values with this value. If both DataFrame locations are missing, the
result will be missing.
df.add(df1, fill_value=0)
Out:
0 1 2 3 4
A 3.0 3.0 2.0 1.0 NaN
B 3.0 3.0 2.0 1.0 NaN
C 3.0 3.0 2.0 1.0 NaN
D 3.0 3.0 2.0 1.0 NaN
E 3.0 3.0 2.0 1.0 NaN