I have a data set like this:
seq S01-T01 S01-T02 S01-T03 S02-T01 S02-T02 S02-T03 S03-T01 S03-T02 S03-T03
B 7 2 9 2 1 9 2 1 1
C NaN 4 4 2 4 NaN 2 6 8
D 5 NaN NaN 2 5 9 NaN 1 1
I want to get a data frame that:
(1) calculates the mean of all the columns with T01 in them
(2) gets the mean per S-number except for T01 (i.e. get the mean of T02 and T03, for each S field)
(3) get the mean of the list of numbers returned from step 2 (i.e. step 2 will return a list of means, one for each S-number, i then want the mean of that list).
So the output for above would be:
T0_means mean_of_other_means
B 3.6 3.83
C 1.3 4.33
D 2.3 2.6
(i just in my head changed the NaNs to 0 for averaging).
I'm getting stuck at the first step, I wrote:
import sys
import pandas as pd
df = pd.read_csv('fat_norm_extracted.csv',sep=',')
list_cols_to_keep = ['S01-T01','S02-T01','S03-T01']
df = df.loc[df['column_name'].isin(list_of_cols_to_keep)]
print(df)
And the error is:
Traceback (most recent call last):
File "calculate_averages.py", line 6, in <module>
df = df.loc[df['column_name'].isin(list_of_cols_to_keep)]
File "/home/slowat/.conda/envs/embedding_nlp/lib/python3.8/site-packages/pandas/core/frame.py", line 3024, in __getitem__
indexer = self.columns.get_loc(key)
File "/home/slowat/.conda/envs/embedding_nlp/lib/python3.8/site-packages/pandas/core/indexes/base.py", line 3082, in get_loc
raise KeyError(key) from err
KeyError: 'column_name'
I know what the error means, that column name is being taken as a string, but not how to fix it. Could someone show me a way around this?
The line
df = df.loc[df['column_name'].isin(list_of_cols_to_keep)]
Is filtering the rows of df where the values of the column named column_name are in the list of value of list_of_cols_to_keep.
If you want to select the columns, you can do :
df = df.loc[:, list_of_cols_to_keep]
Where : is for all rows.
Otherwise you can also use :
df = df.filter(list_of_cols_to_keep)
You can use str.contains to flag the column names that includes T01 via boolean mask msk. Then filter the columns using loc and find mean across columns for T01_means. For mean_of_other_means, you can use msk by using groupby.cumsum on it to create groups; then use groupby.mean across columns to find group means; then use mean yet again to find the mean of means:
df = df.set_index('seq').fillna(0)
msk = df.columns.str.contains('T01')
df['T0_means'] = df.loc[:, msk].mean(axis=1)
df['mean_of_other_means'] = df.drop(columns='T0_means').loc[:, ~msk].groupby(msk.cumsum()[~msk], axis=1).mean().mean(axis=1)
df = df.reset_index()
Output:
seq S01-T01 S01-T02 S01-T03 S02-T01 S02-T02 S02-T03 S03-T01 S03-T02 S03-T03 T0_means mean_of_other_means
0 B 7.0 2.0 9.0 2 1 9.0 2.0 1 1 3.666667 3.833333
1 C 0.0 4.0 4.0 2 4 0.0 2.0 6 8 1.333333 4.333333
2 D 5.0 0.0 0.0 2 5 9.0 0.0 1 1 2.333333 2.666667
First , to me it seems like mean_of_means is nothing but mean of all columns that don't end in T01 because consider row B:
S01-T02 S02-T02 S03-T02 mean
B 2 1 9 (2+1+9)/3
S01-T03 S02-T03 S03-T03 mean
B 9 9 1 (9+9+1/3)
Then the mean of above two means is : ( (2+1+9)/3 + (9+9+1)/3 ) / 2 = (2+1+9+9+9+1)/6
which is nothing but the mean of all columns that don't end in T01!
With that I think you can do:
df = df.fillna(0)
T01_means = df.filter(regex='.*T01$',axis=1).mean(axis=1)
mean_of_means_no_T01 = df.filter(regex='.*(?<!T01)$',axis=1).mean(axis=1)
and then
means_df = pd.concat([T01_means, mean_of_means_no_T01],axis=1)
means_df.columns = ['T01_means', 'mean_of_means_no_T01']
means_df
T01_means mean_of_means_no_T01
B 3.666667 3.833333
C 1.333333 4.333333
D 2.333333 2.666667
Related
I have the following pandas dataframe and would like to build a new column 'c' which is the summation of column 'b' value and column 'a' previous values. With shifting column 'a' it is possible to do so. However, I would like to know how I can pass the previous values of column 'a' in the apply() function.
l1 = [1,2,3,4,5]
l2 = [3,2,5,4,6]
df = pd.DataFrame(data=l1, columns=['a'])
df['b'] = l2
df['shifted'] = df['a'].shift(1)
df['c'] = df.apply(lambda row: row['shifted']+ row['b'], axis=1)
print(df)
a b shifted c
0 1 3 NaN NaN
1 2 2 1.0 3.0
2 3 5 2.0 7.0
3 4 4 3.0 7.0
4 5 6 4.0 10.0
I appreciate your help.
Edit: this is a dummy example. I need to use the apply function because I'm passing another function to it which uses previous rows of some columns and checks some condition.
First let's make it clear that you do not need apply for this simple operation, so I'll consider it as a dummy example of a complex function.
Assuming non-duplicate indices, you can generate a shifted Series and reference it in apply using the name attribute:
s = df['a'].shift(1)
df['c'] =df.apply(lambda row: row['b']+s[row.name], axis=1)
output:
a b shifted c
0 1 3 NaN NaN
1 2 2 1.0 3.0
2 3 5 2.0 7.0
3 4 4 3.0 7.0
4 5 6 4.0 10.0
Code:
data['rolling_sum'] = data.groupby(['User_id'])['Amount'].rolling().sum()
Error
TypeError: incompatible index of inserted column with frame index
Please help in figuring out the mistake in the code. An alternative method would also be appreciated.
Use DataFrame.reset_index with level=0 and drop=True for remove first level of MultiIndex, what is safer because aligned by original index values:
data = pd.DataFrame({
'Amount':[5,3,6,9,2,4],
'User_id':list('aababb')
})
data['rolling_sum1'] = data.groupby(['User_id'])['Amount'].rolling(2).sum().reset_index(level=0, drop=True)
If assign only numpy array is possible values are added incorrectly:
data['rolling_sum2'] = data.groupby(['User_id'])['Amount'].rolling(2).sum().values
print (data)
Amount User_id rolling_sum1 rolling_sum2
0 5 a NaN NaN
1 3 a 8.0 8.0
2 6 b NaN 12.0
3 9 a 12.0 NaN
4 2 b 8.0 8.0
5 4 b 6.0 6.0
I was wondering if there is an efficient way to add rows to a Dataframe that e.g. include the average or a predifined value in case there are not enough rows for a specific value in another column. I guess the description of the Problem is not the best that is why you find an example below:
Say we have the Dataframe
df1
Client NumberOfProducts ID
A 1 2
A 5 1
B 1 2
B 6 1
C 9 1
And we want to have 2 Rows for each client A, B, C, D, no matter if these 2 rows are already existing or not. So for Client A and B we can just copy the rows, for C we want to add a row which says Client = C, NumberOfProducts = average of existing rows = 9 and ID is not of interest (so we could set it to ID = smallest existing one - 1 = 0 any other value, even NaN, would also be possible). For Client D there does not exist a single row so we want to add 2 rows where NumberOfProducts is equal to the constant 2.5. The output should then look like this:
df1
Client NumberOfProducts ID
A 1 2
A 5 1
B 1 2
B 6 1
C 9 1
C 9 0
D 2.5 NaN
D 2.5 NaN
What I have done so far is to loop through the dataframe and add rows where necessary. Since this is pretty inefficient any better solution would be highly appreciated.
Use:
clients = ['A','B','C','D']
N = 2
#test only values from list and also filter only 2 rows for each client if necessary
df = df[df['Client'].isin(clients)].groupby('Client').head(N)
#create helper counter and reshape by unstack
df1 = df.set_index(['Client',df.groupby('Client').cumcount()]).unstack()
#set first if only 1 row per client - replace second NumberOfProducts by first
df1[('NumberOfProducts',1)] = df1[('NumberOfProducts',1)].fillna(df1[('NumberOfProducts',0)])
# ... replace second ID by first subtracted by 1
df1[('ID',1)] = df1[('ID',1)].fillna(df1[('ID',0)] - 1)
#add missing clients by reindex
df1 = df1.reindex(clients)
#replace NumberOfProducts by constant 2.5
df1['NumberOfProducts'] = df1['NumberOfProducts'].fillna(2.5)
print (df1)
NumberOfProducts ID
0 1 0 1
Client
A 1.0 5.0 2.0 1.0
B 1.0 6.0 2.0 1.0
C 9.0 9.0 1.0 0.0
D 2.5 2.5 NaN NaN
#last reshape to original
df2 = df1.stack().reset_index(level=1, drop=True).reset_index()
print (df2)
Client NumberOfProducts ID
0 A 1.0 2.0
1 A 5.0 1.0
2 B 1.0 2.0
3 B 6.0 1.0
4 C 9.0 1.0
5 C 9.0 0.0
6 D 2.5 NaN
7 D 2.5 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.