I would like to fill N/A values in a DataFrame in a selective manner. In particular, if there is a sequence of consequetive nans within a column, I want them to be filled by the preceeding non-nan value, but only if the length of the nan sequence is below a specified threshold. For example, if the threshold is 3 then a within-column sequence of 3 or less will be filled with the preceeding non-nan value, whereas a sequence of 4 or more nans will be left as is.
That is, if the input DataFrame is
2 5 4
nan nan nan
nan nan nan
5 nan nan
9 3 nan
7 9 1
I want the output to be:
2 5 4
2 5 nan
2 5 nan
5 5 nan
9 3 nan
7 9 1
The fillna function, when applied to a DataFrame, has the method and limit options. But these are unfortunately not sufficient to acheive the task. I tried to specify method='ffill' and limit=3, but that fills in the first 3 nans of any sequence, not selectively as described above.
I suppose this can be coded by going column by column with some conditional statements, but I suspect there must be something more Pythonic. Any suggestinos on an efficient way to acheive this?
Working with contiguous groups is still a little awkward in pandas.. or at least I don't know of a slick way to do this, which isn't at all the same thing. :-)
One way to get what you want would be to use the compare-cumsum-groupby pattern:
In [68]: nulls = df.isnull()
...: groups = (nulls != nulls.shift()).cumsum()
...: to_fill = groups.apply(lambda x: x.groupby(x).transform(len) <= 3)
...: df.where(~to_fill, df.ffill())
...:
Out[68]:
0 1 2
0 2.0 5.0 4.0
1 2.0 5.0 NaN
2 2.0 5.0 NaN
3 5.0 5.0 NaN
4 9.0 3.0 NaN
5 7.0 9.0 1.0
Okay, another alternative which I don't like because it's too tricky:
def method_2(df):
nulls = df.isnull()
filled = df.ffill(limit=3)
unfilled = nulls & (~filled.notnull())
nf = nulls.replace({False: 2.0, True: np.nan})
do_not_fill = nf.combine_first(unfilled.replace(False, np.nan)).bfill() == 1
return df.where(do_not_fill, df.ffill())
This doesn't use any groupby tools and so should be faster. Note that a different approach would be to manually (using shifts) determine which elements are to be filled because they're a group of length 1, 2, or 3.
Related
I have a dataframe with around 50 columns and around 3000 rows. Most cells are empty but not all of them. I am trying to add a new row at the end of the dataframe, with the mean value of each column and I need it to ignore the empty cells.
I am using df.mean(axis=0), which somehows turns all values of the dataframe into imaginary numbers. All values stay the same but a +0j is added. I have no Idea why.
Turbine.loc['Mean_Values'] = Turbine.mean(axis=0)
I couldnt find a solution for this, is it because of the empty cells?
Base on this, df.mean() will automatically skip the NaN/Null value with parameter value of skipna=True. Example:
df=pd.DataFrame({'value':[1,2,3,np.nan,5,6,np.nan]})
df=df.append({'value':df.mean(numeric_only=True).value}, ignore_index=True,)
print(df)
Output:
value
0 1.0
1 2.0
2 3.0
3 NaN
4 5.0
5 6.0
6 NaN
7 3.4
But if there is a complex number in a cell, the result of df.mean() will be cast to complex number. Example:
df=pd.DataFrame({'value':[1,2,3,np.nan,5,6,np.nan, complex(1,0)]})
print(df)
print('\n')
df=df.append({'value':df.mean(numeric_only=True).value}, ignore_index=True,)
print(df)
Output with a complex value in a cell:
value
0 (1+0j)
1 (2+0j)
2 (3+0j)
3 NaN
4 (5+0j)
5 (6+0j)
6 NaN
7 (1+0j)
value
0 (1+0j)
1 (2+0j)
2 (3+0j)
3 NaN
4 (5+0j)
5 (6+0j)
6 NaN
7 (1+0j)
8 (3+0j)
Hope this can help you :)
some cells had information about directions (north, west...) in them, which were interpreted as imaginary numbers.
I follow the method in this post to replace missing values with the group mode, but encounter the "IndexError: index out of bounds".
df['SIC'] = df.groupby('CIK').SIC.apply(lambda x: x.fillna(x.mode()[0]))
I guess this is probably because some groups have all missing values and do not have a mode. Is there a way to get around this? Thank you!
mode is quite difficult, given that there really isn't any agreed upon way to deal with ties. Plus it's typically very slow. Here's one way that will be "fast". We'll define a function that calculates the mode for each group, then we can fill the missing values afterwards with a map. We don't run into issues with missing groups, though for ties we arbitrarily choose the modal value that comes first when sorted:
def fast_mode(df, key_cols, value_col):
"""
Calculate a column mode, by group, ignoring null values.
Parameters
----------
df : pandas.DataFrame
DataFrame over which to calcualate the mode.
key_cols : list of str
Columns to groupby for calculation of mode.
value_col : str
Column for which to calculate the mode.
Return
------
pandas.DataFrame
One row for the mode of value_col per key_cols group. If ties,
returns the one which is sorted first.
"""
return (df.groupby(key_cols + [value_col]).size()
.to_frame('counts').reset_index()
.sort_values('counts', ascending=False)
.drop_duplicates(subset=key_cols)).drop(columns='counts')
Sample data df:
CIK SIK
0 C 2.0
1 C 1.0
2 B NaN
3 B 3.0
4 A NaN
5 A 3.0
6 C NaN
7 B NaN
8 C 1.0
9 A 2.0
10 D NaN
11 D NaN
12 D NaN
Code:
df.loc[df.SIK.isnull(), 'SIK'] = df.CIK.map(fast_mode(df, ['CIK'], 'SIK').set_index('CIK').SIK)
Output df:
CIK SIK
0 C 2.0
1 C 1.0
2 B 3.0
3 B 3.0
4 A 2.0
5 A 3.0
6 C 1.0
7 B 3.0
8 C 1.0
9 A 2.0
10 D NaN
11 D NaN
12 D NaN
Given a pandas dataframe containing possible NaN values scattered here and there:
Question: How do I determine which columns contain NaN values? In particular, can I get a list of the column names containing NaNs?
UPDATE: using Pandas 0.22.0
Newer Pandas versions have new methods 'DataFrame.isna()' and 'DataFrame.notna()'
In [71]: df
Out[71]:
a b c
0 NaN 7.0 0
1 0.0 NaN 4
2 2.0 NaN 4
3 1.0 7.0 0
4 1.0 3.0 9
5 7.0 4.0 9
6 2.0 6.0 9
7 9.0 6.0 4
8 3.0 0.0 9
9 9.0 0.0 1
In [72]: df.isna().any()
Out[72]:
a True
b True
c False
dtype: bool
as list of columns:
In [74]: df.columns[df.isna().any()].tolist()
Out[74]: ['a', 'b']
to select those columns (containing at least one NaN value):
In [73]: df.loc[:, df.isna().any()]
Out[73]:
a b
0 NaN 7.0
1 0.0 NaN
2 2.0 NaN
3 1.0 7.0
4 1.0 3.0
5 7.0 4.0
6 2.0 6.0
7 9.0 6.0
8 3.0 0.0
9 9.0 0.0
OLD answer:
Try to use isnull():
In [97]: df
Out[97]:
a b c
0 NaN 7.0 0
1 0.0 NaN 4
2 2.0 NaN 4
3 1.0 7.0 0
4 1.0 3.0 9
5 7.0 4.0 9
6 2.0 6.0 9
7 9.0 6.0 4
8 3.0 0.0 9
9 9.0 0.0 1
In [98]: pd.isnull(df).sum() > 0
Out[98]:
a True
b True
c False
dtype: bool
or as #root proposed clearer version:
In [5]: df.isnull().any()
Out[5]:
a True
b True
c False
dtype: bool
In [7]: df.columns[df.isnull().any()].tolist()
Out[7]: ['a', 'b']
to select a subset - all columns containing at least one NaN value:
In [31]: df.loc[:, df.isnull().any()]
Out[31]:
a b
0 NaN 7.0
1 0.0 NaN
2 2.0 NaN
3 1.0 7.0
4 1.0 3.0
5 7.0 4.0
6 2.0 6.0
7 9.0 6.0
8 3.0 0.0
9 9.0 0.0
You can use df.isnull().sum(). It shows all columns and the total NaNs of each feature.
I had a problem where I had to many columns to visually inspect on the screen so a shortlist comp that filters and returns the offending columns is
nan_cols = [i for i in df.columns if df[i].isnull().any()]
if that's helpful to anyone
Adding to that if you want to filter out columns having more nan values than a threshold, say 85% then use
nan_cols85 = [i for i in df.columns if df[i].isnull().sum() > 0.85*len(data)]
This worked for me,
1. For getting Columns having at least 1 null value. (column names)
data.columns[data.isnull().any()]
2. For getting Columns with count, with having at least 1 null value.
data[data.columns[data.isnull().any()]].isnull().sum()
[Optional]
3. For getting percentage of the null count.
data[data.columns[data.isnull().any()]].isnull().sum() * 100 / data.shape[0]
In datasets having large number of columns its even better to see how many columns contain null values and how many don't.
print("No. of columns containing null values")
print(len(df.columns[df.isna().any()]))
print("No. of columns not containing null values")
print(len(df.columns[df.notna().all()]))
print("Total no. of columns in the dataframe")
print(len(df.columns))
For example in my dataframe it contained 82 columns, of which 19 contained at least one null value.
Further you can also automatically remove cols and rows depending on which has more null values
Here is the code which does this intelligently:
df = df.drop(df.columns[df.isna().sum()>len(df.columns)],axis = 1)
df = df.dropna(axis = 0).reset_index(drop=True)
Note: Above code removes all of your null values. If you want null values, process them before.
df.columns[df.isnull().any()].tolist()
it will return name of columns that contains null rows
I know this is a very well-answered question but I wanted to add a slight adjustment. This answer only returns columns containing nulls, and also still shows the count of the nulls.
As 1-liner:
pd.isnull(df).sum()[pd.isnull(df).sum() > 0]
Description
Count nulls in each column
null_count_ser = pd.isnull(df).sum()
True|False series describing if that column had nulls
is_null_ser = null_count_ser > 0
Use the T|F series to filter out those without
null_count_ser[is_null_ser]
Example Output
name 5
phone 187
age 644
i use these three lines of code to print out the column names which contain at least one null value:
for column in dataframe:
if dataframe[column].isnull().any():
print('{0} has {1} null values'.format(column, dataframe[column].isnull().sum()))
This is one of the methods..
import pandas as pd
df = pd.DataFrame({'a':[1,2,np.nan], 'b':[np.nan,1,np.nan],'c':[np.nan,2,np.nan], 'd':[np.nan,np.nan,np.nan]})
print(pd.isnull(df).sum())
enter image description here
Both of these should work:
df.isnull().sum()
df.isna().sum()
DataFrame methods isna() or isnull() are completely identical.
Note: Empty strings '' is considered as False (not considered NA)
df.isna() return True values for NaN, False for the rest. So, doing:
df.isna().any()
will return True for any column having a NaN, False for the rest
To see just the columns containing NaNs and just the rows containing NaNs:
isnulldf = df.isnull()
columns_containing_nulls = isnulldf.columns[isnulldf.any()]
rows_containing_nulls = df[isnulldf[columns_containing_nulls].any(axis='columns')].index
only_nulls_df = df[columns_containing_nulls].loc[rows_containing_nulls]
print(only_nulls_df)
features_with_na=[features for features in dataframe.columns if dataframe[features].isnull().sum()>0]
for feature in features_with_na:
print(feature, np.round(dataframe[feature].isnull().mean(), 4), '% missing values')
print(features_with_na)
it will give % of missing value for each column in dataframe
The code works if you want to find columns containing NaN values and get a list of the column names.
na_names = df.isnull().any()
list(na_names.where(na_names == True).dropna().index)
If you want to find columns whose values are all NaNs, you can replace any with all.
Given a pandas dataframe containing possible NaN values scattered here and there:
Question: How do I determine which columns contain NaN values? In particular, can I get a list of the column names containing NaNs?
UPDATE: using Pandas 0.22.0
Newer Pandas versions have new methods 'DataFrame.isna()' and 'DataFrame.notna()'
In [71]: df
Out[71]:
a b c
0 NaN 7.0 0
1 0.0 NaN 4
2 2.0 NaN 4
3 1.0 7.0 0
4 1.0 3.0 9
5 7.0 4.0 9
6 2.0 6.0 9
7 9.0 6.0 4
8 3.0 0.0 9
9 9.0 0.0 1
In [72]: df.isna().any()
Out[72]:
a True
b True
c False
dtype: bool
as list of columns:
In [74]: df.columns[df.isna().any()].tolist()
Out[74]: ['a', 'b']
to select those columns (containing at least one NaN value):
In [73]: df.loc[:, df.isna().any()]
Out[73]:
a b
0 NaN 7.0
1 0.0 NaN
2 2.0 NaN
3 1.0 7.0
4 1.0 3.0
5 7.0 4.0
6 2.0 6.0
7 9.0 6.0
8 3.0 0.0
9 9.0 0.0
OLD answer:
Try to use isnull():
In [97]: df
Out[97]:
a b c
0 NaN 7.0 0
1 0.0 NaN 4
2 2.0 NaN 4
3 1.0 7.0 0
4 1.0 3.0 9
5 7.0 4.0 9
6 2.0 6.0 9
7 9.0 6.0 4
8 3.0 0.0 9
9 9.0 0.0 1
In [98]: pd.isnull(df).sum() > 0
Out[98]:
a True
b True
c False
dtype: bool
or as #root proposed clearer version:
In [5]: df.isnull().any()
Out[5]:
a True
b True
c False
dtype: bool
In [7]: df.columns[df.isnull().any()].tolist()
Out[7]: ['a', 'b']
to select a subset - all columns containing at least one NaN value:
In [31]: df.loc[:, df.isnull().any()]
Out[31]:
a b
0 NaN 7.0
1 0.0 NaN
2 2.0 NaN
3 1.0 7.0
4 1.0 3.0
5 7.0 4.0
6 2.0 6.0
7 9.0 6.0
8 3.0 0.0
9 9.0 0.0
You can use df.isnull().sum(). It shows all columns and the total NaNs of each feature.
I had a problem where I had to many columns to visually inspect on the screen so a shortlist comp that filters and returns the offending columns is
nan_cols = [i for i in df.columns if df[i].isnull().any()]
if that's helpful to anyone
Adding to that if you want to filter out columns having more nan values than a threshold, say 85% then use
nan_cols85 = [i for i in df.columns if df[i].isnull().sum() > 0.85*len(data)]
This worked for me,
1. For getting Columns having at least 1 null value. (column names)
data.columns[data.isnull().any()]
2. For getting Columns with count, with having at least 1 null value.
data[data.columns[data.isnull().any()]].isnull().sum()
[Optional]
3. For getting percentage of the null count.
data[data.columns[data.isnull().any()]].isnull().sum() * 100 / data.shape[0]
In datasets having large number of columns its even better to see how many columns contain null values and how many don't.
print("No. of columns containing null values")
print(len(df.columns[df.isna().any()]))
print("No. of columns not containing null values")
print(len(df.columns[df.notna().all()]))
print("Total no. of columns in the dataframe")
print(len(df.columns))
For example in my dataframe it contained 82 columns, of which 19 contained at least one null value.
Further you can also automatically remove cols and rows depending on which has more null values
Here is the code which does this intelligently:
df = df.drop(df.columns[df.isna().sum()>len(df.columns)],axis = 1)
df = df.dropna(axis = 0).reset_index(drop=True)
Note: Above code removes all of your null values. If you want null values, process them before.
df.columns[df.isnull().any()].tolist()
it will return name of columns that contains null rows
I know this is a very well-answered question but I wanted to add a slight adjustment. This answer only returns columns containing nulls, and also still shows the count of the nulls.
As 1-liner:
pd.isnull(df).sum()[pd.isnull(df).sum() > 0]
Description
Count nulls in each column
null_count_ser = pd.isnull(df).sum()
True|False series describing if that column had nulls
is_null_ser = null_count_ser > 0
Use the T|F series to filter out those without
null_count_ser[is_null_ser]
Example Output
name 5
phone 187
age 644
i use these three lines of code to print out the column names which contain at least one null value:
for column in dataframe:
if dataframe[column].isnull().any():
print('{0} has {1} null values'.format(column, dataframe[column].isnull().sum()))
This is one of the methods..
import pandas as pd
df = pd.DataFrame({'a':[1,2,np.nan], 'b':[np.nan,1,np.nan],'c':[np.nan,2,np.nan], 'd':[np.nan,np.nan,np.nan]})
print(pd.isnull(df).sum())
enter image description here
Both of these should work:
df.isnull().sum()
df.isna().sum()
DataFrame methods isna() or isnull() are completely identical.
Note: Empty strings '' is considered as False (not considered NA)
df.isna() return True values for NaN, False for the rest. So, doing:
df.isna().any()
will return True for any column having a NaN, False for the rest
To see just the columns containing NaNs and just the rows containing NaNs:
isnulldf = df.isnull()
columns_containing_nulls = isnulldf.columns[isnulldf.any()]
rows_containing_nulls = df[isnulldf[columns_containing_nulls].any(axis='columns')].index
only_nulls_df = df[columns_containing_nulls].loc[rows_containing_nulls]
print(only_nulls_df)
features_with_na=[features for features in dataframe.columns if dataframe[features].isnull().sum()>0]
for feature in features_with_na:
print(feature, np.round(dataframe[feature].isnull().mean(), 4), '% missing values')
print(features_with_na)
it will give % of missing value for each column in dataframe
The code works if you want to find columns containing NaN values and get a list of the column names.
na_names = df.isnull().any()
list(na_names.where(na_names == True).dropna().index)
If you want to find columns whose values are all NaNs, you can replace any with all.
I am trying to do the following: on a dataframe X, I want to select all rows where X['a']>0 but I want to preserve the dimension of X, so that any other row will appear as containing NaN. Is there a fast way to do it? If one does X[X['a']>0] the dimensions of X are not preserved.
Use double subscript [[]]:
In [42]:
df = pd.DataFrame({'a':np.random.randn(10)})
df
Out[42]:
a
0 1.042971
1 0.978914
2 0.764374
3 -0.338405
4 0.974011
5 -0.995945
6 -1.649612
7 0.965838
8 -0.142608
9 -0.804508
In [48]:
df[df[['a']] > 1]
Out[48]:
a
0 1.042971
1 NaN
2 NaN
3 NaN
4 NaN
5 NaN
6 NaN
7 NaN
8 NaN
9 NaN
The key semantic difference here is what is returned is a df when you double subscript so this masks the df itself rather than the index
Note though that if you have multiple columns then it will mask all those as NaN