None in if condition, how to handle missing data? - python

If the value of age is missing I want to create a variable with the value of 1. Instead everything is None in the output of the Value column.
raw_data1 = {'id': [1,2,3,5],
'age': [0, np.nan, 10, 2]}
df1 = pd.DataFrame(raw_data1, columns = ['id','age'])
def my_test(b):
if b is None:
return 1
df1['Value'] = df1.apply(lambda row: my_test(row['age']), axis=1)
How can implement it? I know that there are several ways, but I would like to focus on the use of a function, (def my_test etc.).


If I understood you correctly, you can use:
df1['value'] = np.where(df1['age'].isnull(), 1, '')
Output:
id age value
0 1 0.0
1 2 NaN 1
2 3 10.0
3 5 2.0


You can use row.get('age') instead of row['age'].
get() returns null if age is not inside the dict


Do this instead,
>>> df1.value = df1.age.isna().astype(int)
>>> df1
id age value
0 1 0.0 0
1 2 NaN 1
2 3 10.0 0
3 5 2.0 0


You can use map for this
df1['Value'] = df1['age'].map(lambda x : 1 if np.isnan(x) else np.nan)
If you want to make use of your function, you can use map like this
def my_test(b):
if np.isnan(b):
return 1
else:
return np.nan
df1['Value'] = df1['age'].map(lambda x : my_test(x))

Related

How to create new column and assign values by column group

I have dataframe,
df
uid
1
2
3
...
I want to assign a new column, with values 0 or 1 depending on the uid, which I will assign.
df
uid new
1 0
2 0
3 1
..

You must explain the underlying logic.
That said, there are many possible ways.
Considering an explicit mapping with map:
mapper = {1: 0, 2: 0, 3: 1}
df['new'] = df['uid'].map(mapper)
# or
mapper = {0: [1, 2], 1: [3]}
df['new'] = df['uid'].map({k:v for v,l in mapper.items() for k in l})
Or using a list of targets for the 1 with isin and conversion to int
target = [3]
df['new'] = df['uid'].isin(target).astype(int)
Output:
uid new
0 1 0
1 2 0
2 3 1

If there is a correlation between uid and new, you can create a function to define the mapping between uid and new
def mapping(value):
new_value = value // 2
return new_value
Then
df["new"] = df["uid"].apply(mapping)
Or directly
df["new"] = df["uid"].apply(lambda value: value // 2)

from the relation of 3 uid's I came up as a relation the uid which is divisible by 3 is assign to 1 else 0. (Not sure the relation is correct or not as you have given only 3 values of uid)
you can apply np.where() -> np.where(condition, x, y) if condition satisfy it assign value x else value y
import pandas as pd
import numpy as np
df = pd.DataFrame({'uid': [1, 2, 3]})
df["new"] = np.where(df["uid"] % 3 == 0, 1, 0)
print(df)
Output:
uid new
0 1 0
1 2 0
2 3 1

If substring in string, replace string with number

I have the following mapping
mapping = {'sum12':2, 'sum6':1,
'avg12':2, 'avg6':1,
'diff':3, 'mean':4}
and I have a dataframe with variables like
var1 var2
0 abc_sum12 mean_jkl
1 pqr_sum6 pqr_avg6
2 diff_xyz qwerty
If any of the substrings are present in the strings in the dataframe, I want to replace them with their corresponding values. If no substring is present, I want to replace it with np.nan. At present, the only solution I can think of is going through every row, checking if any of the substrings is present in every string, and replacing it with the specific number corresponding with that substring. Is there a better way to do it.
The output in the end would be
var1 var2
0 2 4.0
1 1 1.0
2 3 NaN

I believe if you replace using regex, it will partial match and give you the result you want. The only exception is the qwerty value which will remain unchanged. If you then coerce the entire df to numeric, it will return NaN for that value or any other non-numeric.
import pandas as pd
mapping = {'sum12':2, 'sum6':1,
'avg12':2, 'avg6':1,
'diff':3, 'mean':4}
df = pd.DataFrame({'var1': ['abc_sum12', 'pqr_sum6', 'diff_xyz'],
'var2': ['mean_jkl', 'pqr_avg6', 'qwerty']})
df = df.replace(mapping, regex=True).apply(lambda x: pd.to_numeric(x, errors='coerce'))
print(df)
output
var1 var2
0 2 4.0
1 1 1.0
2 3 NaN

Another approach:
mapping = {'sum12':2, 'sum6':1,
'avg12':2, 'avg6':1,
'diff':3, 'mean':4}
df = pd.DataFrame(
{'var1': {0: 'abc_sum12', 1: 'pqr_sum6', 2: 'diff_xyz'},
'var2': {0: 'mean_jkl', 1: 'pqr_avg6', 2: 'qwerty'}})
df_new[:] = np.nan
df_new = df_new.astype('float')
for name,col in df.items():
for key,val in mapping.items():
df_new[name][col.str.contains(key)] = val
The resulting dataframe df_new:
var1 var2
0 2.0 4.0
1 1.0 1.0
2 3.0 NaN

Fill NA values in dataframe by mapping a double indexed groupby object [duplicate]

This should be straightforward, but the closest thing I've found is this post:
pandas: Filling missing values within a group, and I still can't solve my problem....
Suppose I have the following dataframe
df = pd.DataFrame({'value': [1, np.nan, np.nan, 2, 3, 1, 3, np.nan, 3], 'name': ['A','A', 'B','B','B','B', 'C','C','C']})
name value
0 A 1
1 A NaN
2 B NaN
3 B 2
4 B 3
5 B 1
6 C 3
7 C NaN
8 C 3
and I'd like to fill in "NaN" with mean value in each "name" group, i.e.
name value
0 A 1
1 A 1
2 B 2
3 B 2
4 B 3
5 B 1
6 C 3
7 C 3
8 C 3
I'm not sure where to go after:
grouped = df.groupby('name').mean()
Thanks a bunch.

One way would be to use transform:
>>> df
name value
0 A 1
1 A NaN
2 B NaN
3 B 2
4 B 3
5 B 1
6 C 3
7 C NaN
8 C 3
>>> df["value"] = df.groupby("name").transform(lambda x: x.fillna(x.mean()))
>>> df
name value
0 A 1
1 A 1
2 B 2
3 B 2
4 B 3
5 B 1
6 C 3
7 C 3
8 C 3

fillna + groupby + transform + mean
This seems intuitive:
df['value'] = df['value'].fillna(df.groupby('name')['value'].transform('mean'))
The groupby + transform syntax maps the groupwise mean to the index of the original dataframe. This is roughly equivalent to #DSM's solution, but avoids the need to define an anonymous lambda function.

#DSM has IMO the right answer, but I'd like to share my generalization and optimization of the question: Multiple columns to group-by and having multiple value columns:
df = pd.DataFrame(
{
'category': ['X', 'X', 'X', 'X', 'X', 'X', 'Y', 'Y', 'Y'],
'name': ['A','A', 'B','B','B','B', 'C','C','C'],
'other_value': [10, np.nan, np.nan, 20, 30, 10, 30, np.nan, 30],
'value': [1, np.nan, np.nan, 2, 3, 1, 3, np.nan, 3],
}
)
... gives ...
category name other_value value
0 X A 10.0 1.0
1 X A NaN NaN
2 X B NaN NaN
3 X B 20.0 2.0
4 X B 30.0 3.0
5 X B 10.0 1.0
6 Y C 30.0 3.0
7 Y C NaN NaN
8 Y C 30.0 3.0
In this generalized case we would like to group by category and name, and impute only on value.
This can be solved as follows:
df['value'] = df.groupby(['category', 'name'])['value']\
.transform(lambda x: x.fillna(x.mean()))
Notice the column list in the group-by clause, and that we select the value column right after the group-by. This makes the transformation only be run on that particular column. You could add it to the end, but then you will run it for all columns only to throw out all but one measure column at the end. A standard SQL query planner might have been able to optimize this, but pandas (0.19.2) doesn't seem to do this.
Performance test by increasing the dataset by doing ...
big_df = None
for _ in range(10000):
if big_df is None:
big_df = df.copy()
else:
big_df = pd.concat([big_df, df])
df = big_df
... confirms that this increases the speed proportional to how many columns you don't have to impute:
import pandas as pd
from datetime import datetime
def generate_data():
...
t = datetime.now()
df = generate_data()
df['value'] = df.groupby(['category', 'name'])['value']\
.transform(lambda x: x.fillna(x.mean()))
print(datetime.now()-t)
# 0:00:00.016012
t = datetime.now()
df = generate_data()
df["value"] = df.groupby(['category', 'name'])\
.transform(lambda x: x.fillna(x.mean()))['value']
print(datetime.now()-t)
# 0:00:00.030022
On a final note you can generalize even further if you want to impute more than one column, but not all:
df[['value', 'other_value']] = df.groupby(['category', 'name'])['value', 'other_value']\
.transform(lambda x: x.fillna(x.mean()))

Shortcut:
Groupby + Apply + Lambda + Fillna + Mean
>>> df['value1']=df.groupby('name')['value'].apply(lambda x:x.fillna(x.mean()))
>>> df.isnull().sum().sum()
0
This solution still works if you want to group by multiple columns to replace missing values.
>>> df = pd.DataFrame({'value': [1, np.nan, np.nan, 2, 3, np.nan,np.nan, 4, 3],
'name': ['A','A', 'B','B','B','B', 'C','C','C'],'class':list('ppqqrrsss')})
>>> df['value']=df.groupby(['name','class'])['value'].apply(lambda x:x.fillna(x.mean()))
>>> df
value name class
0 1.0 A p
1 1.0 A p
2 2.0 B q
3 2.0 B q
4 3.0 B r
5 3.0 B r
6 3.5 C s
7 4.0 C s
8 3.0 C s

I'd do it this way
df.loc[df.value.isnull(), 'value'] = df.groupby('group').value.transform('mean')

The featured high ranked answer only works for a pandas Dataframe with only two columns. If you have a more columns case use instead:
df['Crude_Birth_rate'] = df.groupby("continent").Crude_Birth_rate.transform(
lambda x: x.fillna(x.mean()))

To summarize all above concerning the efficiency of the possible solution
I have a dataset with 97 906 rows and 48 columns.
I want to fill in 4 columns with the median of each group.
The column I want to group has 26 200 groups.
The first solution
start = time.time()
x = df_merged[continuous_variables].fillna(df_merged.groupby('domain_userid')[continuous_variables].transform('median'))
print(time.time() - start)
0.10429811477661133 seconds
The second solution
start = time.time()
for col in continuous_variables:
df_merged.loc[df_merged[col].isnull(), col] = df_merged.groupby('domain_userid')[col].transform('median')
print(time.time() - start)
0.5098445415496826 seconds
The next solution I only performed on a subset since it was running too long.
start = time.time()
for col in continuous_variables:
x = df_merged.head(10000).groupby('domain_userid')[col].transform(lambda x: x.fillna(x.median()))
print(time.time() - start)
11.685635566711426 seconds
The following solution follows the same logic as above.
start = time.time()
x = df_merged.head(10000).groupby('domain_userid')[continuous_variables].transform(lambda x: x.fillna(x.median()))
print(time.time() - start)
42.630549907684326 seconds
So it's quite important to choose the right method.
Bear in mind that I noticed once a column was not a numeric the times were going up exponentially (makes sense as I was computing the median).

def groupMeanValue(group):
group['value'] = group['value'].fillna(group['value'].mean())
return group
dft = df.groupby("name").transform(groupMeanValue)

I know that is an old question. But I am quite surprised by the unanimity of apply/lambda answers here.
Generally speaking, that is the second worst thing to do after iterating rows, from timing point of view.
What I would do here is
df.loc[df['value'].isna(), 'value'] = df.groupby('name')['value'].transform('mean')
Or using fillna
df['value'] = df['value'].fillna(df.groupby('name')['value'].transform('mean'))
I've checked with timeit (because, again, unanimity for apply/lambda based solution made me doubt my instinct). And that is indeed 2.5 faster than the most upvoted solutions.

To fill all the numeric null values with the mean grouped by "name"
num_cols = df.select_dtypes(exclude='object').columns
df[num_cols] = df.groupby("name").transform(lambda x: x.fillna(x.mean()))

df.fillna(df.groupby(['name'], as_index=False).mean(), inplace=True)

You can also use "dataframe or table_name".apply(lambda x: x.fillna(x.mean())).

Fillna in pandas with respect to similar line [duplicate]

This should be straightforward, but the closest thing I've found is this post:
pandas: Filling missing values within a group, and I still can't solve my problem....
Suppose I have the following dataframe
df = pd.DataFrame({'value': [1, np.nan, np.nan, 2, 3, 1, 3, np.nan, 3], 'name': ['A','A', 'B','B','B','B', 'C','C','C']})
name value
0 A 1
1 A NaN
2 B NaN
3 B 2
4 B 3
5 B 1
6 C 3
7 C NaN
8 C 3
and I'd like to fill in "NaN" with mean value in each "name" group, i.e.
name value
0 A 1
1 A 1
2 B 2
3 B 2
4 B 3
5 B 1
6 C 3
7 C 3
8 C 3
I'm not sure where to go after:
grouped = df.groupby('name').mean()
Thanks a bunch.

One way would be to use transform:
>>> df
name value
0 A 1
1 A NaN
2 B NaN
3 B 2
4 B 3
5 B 1
6 C 3
7 C NaN
8 C 3
>>> df["value"] = df.groupby("name").transform(lambda x: x.fillna(x.mean()))
>>> df
name value
0 A 1
1 A 1
2 B 2
3 B 2
4 B 3
5 B 1
6 C 3
7 C 3
8 C 3

fillna + groupby + transform + mean
This seems intuitive:
df['value'] = df['value'].fillna(df.groupby('name')['value'].transform('mean'))
The groupby + transform syntax maps the groupwise mean to the index of the original dataframe. This is roughly equivalent to #DSM's solution, but avoids the need to define an anonymous lambda function.

#DSM has IMO the right answer, but I'd like to share my generalization and optimization of the question: Multiple columns to group-by and having multiple value columns:
df = pd.DataFrame(
{
'category': ['X', 'X', 'X', 'X', 'X', 'X', 'Y', 'Y', 'Y'],
'name': ['A','A', 'B','B','B','B', 'C','C','C'],
'other_value': [10, np.nan, np.nan, 20, 30, 10, 30, np.nan, 30],
'value': [1, np.nan, np.nan, 2, 3, 1, 3, np.nan, 3],
}
)
... gives ...
category name other_value value
0 X A 10.0 1.0
1 X A NaN NaN
2 X B NaN NaN
3 X B 20.0 2.0
4 X B 30.0 3.0
5 X B 10.0 1.0
6 Y C 30.0 3.0
7 Y C NaN NaN
8 Y C 30.0 3.0
In this generalized case we would like to group by category and name, and impute only on value.
This can be solved as follows:
df['value'] = df.groupby(['category', 'name'])['value']\
.transform(lambda x: x.fillna(x.mean()))
Notice the column list in the group-by clause, and that we select the value column right after the group-by. This makes the transformation only be run on that particular column. You could add it to the end, but then you will run it for all columns only to throw out all but one measure column at the end. A standard SQL query planner might have been able to optimize this, but pandas (0.19.2) doesn't seem to do this.
Performance test by increasing the dataset by doing ...
big_df = None
for _ in range(10000):
if big_df is None:
big_df = df.copy()
else:
big_df = pd.concat([big_df, df])
df = big_df
... confirms that this increases the speed proportional to how many columns you don't have to impute:
import pandas as pd
from datetime import datetime
def generate_data():
...
t = datetime.now()
df = generate_data()
df['value'] = df.groupby(['category', 'name'])['value']\
.transform(lambda x: x.fillna(x.mean()))
print(datetime.now()-t)
# 0:00:00.016012
t = datetime.now()
df = generate_data()
df["value"] = df.groupby(['category', 'name'])\
.transform(lambda x: x.fillna(x.mean()))['value']
print(datetime.now()-t)
# 0:00:00.030022
On a final note you can generalize even further if you want to impute more than one column, but not all:
df[['value', 'other_value']] = df.groupby(['category', 'name'])['value', 'other_value']\
.transform(lambda x: x.fillna(x.mean()))

Shortcut:
Groupby + Apply + Lambda + Fillna + Mean
>>> df['value1']=df.groupby('name')['value'].apply(lambda x:x.fillna(x.mean()))
>>> df.isnull().sum().sum()
0
This solution still works if you want to group by multiple columns to replace missing values.
>>> df = pd.DataFrame({'value': [1, np.nan, np.nan, 2, 3, np.nan,np.nan, 4, 3],
'name': ['A','A', 'B','B','B','B', 'C','C','C'],'class':list('ppqqrrsss')})
>>> df['value']=df.groupby(['name','class'])['value'].apply(lambda x:x.fillna(x.mean()))
>>> df
value name class
0 1.0 A p
1 1.0 A p
2 2.0 B q
3 2.0 B q
4 3.0 B r
5 3.0 B r
6 3.5 C s
7 4.0 C s
8 3.0 C s

I'd do it this way
df.loc[df.value.isnull(), 'value'] = df.groupby('group').value.transform('mean')

The featured high ranked answer only works for a pandas Dataframe with only two columns. If you have a more columns case use instead:
df['Crude_Birth_rate'] = df.groupby("continent").Crude_Birth_rate.transform(
lambda x: x.fillna(x.mean()))

To summarize all above concerning the efficiency of the possible solution
I have a dataset with 97 906 rows and 48 columns.
I want to fill in 4 columns with the median of each group.
The column I want to group has 26 200 groups.
The first solution
start = time.time()
x = df_merged[continuous_variables].fillna(df_merged.groupby('domain_userid')[continuous_variables].transform('median'))
print(time.time() - start)
0.10429811477661133 seconds
The second solution
start = time.time()
for col in continuous_variables:
df_merged.loc[df_merged[col].isnull(), col] = df_merged.groupby('domain_userid')[col].transform('median')
print(time.time() - start)
0.5098445415496826 seconds
The next solution I only performed on a subset since it was running too long.
start = time.time()
for col in continuous_variables:
x = df_merged.head(10000).groupby('domain_userid')[col].transform(lambda x: x.fillna(x.median()))
print(time.time() - start)
11.685635566711426 seconds
The following solution follows the same logic as above.
start = time.time()
x = df_merged.head(10000).groupby('domain_userid')[continuous_variables].transform(lambda x: x.fillna(x.median()))
print(time.time() - start)
42.630549907684326 seconds
So it's quite important to choose the right method.
Bear in mind that I noticed once a column was not a numeric the times were going up exponentially (makes sense as I was computing the median).

def groupMeanValue(group):
group['value'] = group['value'].fillna(group['value'].mean())
return group
dft = df.groupby("name").transform(groupMeanValue)

I know that is an old question. But I am quite surprised by the unanimity of apply/lambda answers here.
Generally speaking, that is the second worst thing to do after iterating rows, from timing point of view.
What I would do here is
df.loc[df['value'].isna(), 'value'] = df.groupby('name')['value'].transform('mean')
Or using fillna
df['value'] = df['value'].fillna(df.groupby('name')['value'].transform('mean'))
I've checked with timeit (because, again, unanimity for apply/lambda based solution made me doubt my instinct). And that is indeed 2.5 faster than the most upvoted solutions.

To fill all the numeric null values with the mean grouped by "name"
num_cols = df.select_dtypes(exclude='object').columns
df[num_cols] = df.groupby("name").transform(lambda x: x.fillna(x.mean()))

df.fillna(df.groupby(['name'], as_index=False).mean(), inplace=True)

You can also use "dataframe or table_name".apply(lambda x: x.fillna(x.mean())).

Update row values where certain condition is met in pandas

Say I have the following dataframe:
What is the most efficient way to update the values of the columns feat and another_feat where the stream is number 2?
Is this it?
for index, row in df.iterrows():
if df1.loc[index,'stream'] == 2:
# do something
How do I do it if there are more than 100 columns? I don't want to explicitly name the columns that I want to update. I want to divide the value of each column by 2 (except for the stream column).
So to be clear, my goal is:
Dividing all values by 2 of all rows that have stream 2, but not changing the stream column.

I think you can use loc if you need update two columns to same value:
df1.loc[df1['stream'] == 2, ['feat','another_feat']] = 'aaaa'
print df1
stream feat another_feat
a 1 some_value some_value
b 2 aaaa aaaa
c 2 aaaa aaaa
d 3 some_value some_value
If you need update separate, one option is use:
df1.loc[df1['stream'] == 2, 'feat'] = 10
print df1
stream feat another_feat
a 1 some_value some_value
b 2 10 some_value
c 2 10 some_value
d 3 some_value some_value
Another common option is use numpy.where:
df1['feat'] = np.where(df1['stream'] == 2, 10,20)
print df1
stream feat another_feat
a 1 20 some_value
b 2 10 some_value
c 2 10 some_value
d 3 20 some_value
EDIT: If you need divide all columns without stream where condition is True, use:
print df1
stream feat another_feat
a 1 4 5
b 2 4 5
c 2 2 9
d 3 1 7
#filter columns all without stream
cols = [col for col in df1.columns if col != 'stream']
print cols
['feat', 'another_feat']
df1.loc[df1['stream'] == 2, cols ] = df1 / 2
print df1
stream feat another_feat
a 1 4.0 5.0
b 2 2.0 2.5
c 2 1.0 4.5
d 3 1.0 7.0
If working with multiple conditions is possible use multiple numpy.where
or numpy.select:
df0 = pd.DataFrame({'Col':[5,0,-6]})
df0['New Col1'] = np.where((df0['Col'] > 0), 'Increasing',
np.where((df0['Col'] < 0), 'Decreasing', 'No Change'))
df0['New Col2'] = np.select([df0['Col'] > 0, df0['Col'] < 0],
['Increasing', 'Decreasing'],
default='No Change')
print (df0)
Col New Col1 New Col2
0 5 Increasing Increasing
1 0 No Change No Change
2 -6 Decreasing Decreasing

You can do the same with .ix, like this:
In [1]: df = pd.DataFrame(np.random.randn(5,4), columns=list('abcd'))
In [2]: df
Out[2]:
a b c d
0 -0.323772 0.839542 0.173414 -1.341793
1 -1.001287 0.676910 0.465536 0.229544
2 0.963484 -0.905302 -0.435821 1.934512
3 0.266113 -0.034305 -0.110272 -0.720599
4 -0.522134 -0.913792 1.862832 0.314315
In [3]: df.ix[df.a>0, ['b','c']] = 0
In [4]: df
Out[4]:
a b c d
0 -0.323772 0.839542 0.173414 -1.341793
1 -1.001287 0.676910 0.465536 0.229544
2 0.963484 0.000000 0.000000 1.934512
3 0.266113 0.000000 0.000000 -0.720599
4 -0.522134 -0.913792 1.862832 0.314315
EDIT
After the extra information, the following will return all columns - where some condition is met - with halved values:
>> condition = df.a > 0
>> df[condition][[i for i in df.columns.values if i not in ['a']]].apply(lambda x: x/2)

Another vectorized solution is to use the mask() method to halve the rows corresponding to stream=2 and join() these columns to a dataframe that consists only of the stream column:
cols = ['feat', 'another_feat']
df[['stream']].join(df[cols].mask(df['stream'] == 2, lambda x: x/2))
or you can also update() the original dataframe:
df.update(df[cols].mask(df['stream'] == 2, lambda x: x/2))
Both of the above codes do the following:
mask() is even simpler to use if the value to replace is a constant (not derived using a function); e.g. the following code replaces all feat values corresponding to stream equal to 1 or 3 by 100.1
df[['stream']].join(df.filter(like='feat').mask(df['stream'].isin([1,3]), 100))
1: feat columns can be selected using filter() method as well.

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