Currently I have the DataFrame seen below and I want to do a rolling average over the last 10 occurrences that have actual values, but to skip the NaNs
Example DataFrame
The issues is that if I run df['AST_Hit'].rolling(10).mean(skipna=True).shift(1) I get this DataFrame below which is not what I am looking for
Example Output DataFrame
I've tried using window and min_period but that does not give me what I want as I don't want the average over anything greater than 10.
Ideally I would like the DataFrame to be able to discard a NaN, but still look to see if there are 10 values in that selection. From what I am describing I think I need some sort of max period where it is equal to 10 as well as the min period equal to 10, but I could not find anything on Pandas documentation for rolling on setting up a max period.
Maybe it would also be best if I just dropped any NaN rows. My DataFrame is much bigger than what is seen, so it isn't just those 3 rows that contain a NaN, but it may be the best course of action
Any help or tips is greatly appreciated.
This might help you:
import pandas as pd
import numpy as np
# create a sample DataFrame with non-numeric columns
np.random.seed(123)
df = pd.DataFrame({
'Date': pd.date_range(start='2022-01-01', periods=100),
'AST_Hit': np.random.randint(0, 10, size=100),
'Other_Column': ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j'] * 10
})
df.iloc[3:6, 1] = np.nan
df.iloc[7, 0] = np.nan
df.iloc[10:15, 2] = np.nan
df.iloc[20:25, 1] = np.nan
df.iloc[30:40, 2] = np.nan
# compute rolling average over the last 10 non-null values
rolling_mask = df['AST_Hit'].notnull().rolling(window=10, min_periods=1).sum().eq(10)
result = df['AST_Hit'].rolling(window=10, min_periods=1).apply(lambda x: np.mean(x[rolling_mask]))
print(result)
which gives
0 NaN
1 NaN
2 NaN
3 NaN
4 NaN
...
95 5.6
96 5.1
97 4.7
98 4.2
99 3.9
Name: AST_Hit, Length: 100, dtype: float64
Related
I have four different datasets. I have merged three of the dataframes correctly. I have same name column in 3rd and 4th dataset. When I merge it with 4th dataset. I am not getting the same name column values in well mannerd way. The user_id is repeating when I merge. I don't want to repeat the user_id. I want to see the value in the del_keys column where it's showing me NaN value rather than it's showing me the value in the last of table. Moreover, I want to merge values of same name column on the basis of their user_id.
In the above image you can see what kind of problem I am getting.
My expected output will look like. There should not be repeated user_id.
using merge on user_id column
import pandas as pd
import numpy as np
df1 = pd.DataFrame({
'user_id': [1, 2, 3, 4],
'del': [1.0, np.nan, np.nan, np.nan]
})
df2 = pd.DataFrame({
'user_id': [3, 4, 5],
'del_keys': [1.0, 2.0, 3.0]
})
final=df.merge(df2,on="user_id",how="outer")
Combine first to get rid of Nan values and then drop duplicates
final["del_keys"]=final['del_keys_y'].combine_first(final['del_keys_x'])
final.drop(columns=["del_keys_x","del_keys_y"],inplace=True)
final.drop_duplicates(subset="user_id")
I'm guessing that you use pd.concat to merge the dataframes.
Some dataframes:
import pandas as pd
import numpy as np
df1 = pd.DataFrame({
'user_id': [1, 2, 3],
'del_keys': [1.0, np.nan, np.nan]
})
df2 = pd.DataFrame({
'user_id': [3, 4, 5],
'del_keys': [1.0, 2.0, 3.0]
})
Merge using pd.concat:
df = pd.concat([df1, df2])
>>> user_id del_keys
0 1 1.0
1 2 NaN
2 3 NaN
0 3 1.0
1 4 2.0
2 5 3.0
Remove duplicates using pd.drop_duplicates:
(
df
.sort_values('del_keys')
.drop_duplicates('user_id', keep='first')
.sort_values('user_id')
)
>>> user_id del_keys
0 1 1.0
1 2 NaN
0 3 1.0
1 4 2.0
2 5 3.0
First, we sort the values by del_keys such that all NaNs are the bottom of the dataframe. Then we can drop the duplicates and keep the first occurrence for each user_id. Lastly, we can sort again to restore the original order.
I am trying to clean a number of columns in a dataset and try to iterate to different columns.
import pandas as pd
df = pd.DataFrame({
'A': [7.3\N\P,nan\T\Z,11.0\R\Z],
'B': [nan\J\N, nan\A\G, 10.8\F\U],
'C': [12.4\A\I, 13.3\H\Z, 8.200000000000001\B\W]})
for name, values in df.iloc[:, 0:3].iteritems():
def myreplace(s):
for char in ['\A','\B','\C','\D','\E','\F','\G','\H','\I',
'\J','\K','\L','\M','\\N','\O','\P','\Q','\R',
'\S','\T','\V','\W','\X','\Y','\Z','\\U']:
s = s.map(lambda x: x.replace(char, ''))
return s
df = df.apply(myreplace)
I get the error: 'float' object has no attribue 'replace'
I could run this part on one column and it works, but I need to run it on several columns so this part does not work as I get an error that 'Dataframe'objec has no attribute 'str'
df_data.str.replace('[\\\|A|B|C|D|E|F|G|H|I|J|K|L|M|N|O|P|Q|R|S|T|U|V|W|X|Y|Z]', '')
I am really new to python pandas dataframe. Will appreciate the help
Given, assuming the goal is to extract numbers from the strings:
A B C
0 7.3\N\P nan\J\N 12.4\A\I
1 nan\T\Z nan\A\G 13.3\H\Z
2 11.0\R\Z 10.8\F\U 8.200000000000001\B\W
Doing:
cols = ['A', 'B', 'C']
for col in cols:
df[col] = df[col].str.extract('(\d*\.\d*)').astype(float)
Output:
A B C
0 7.3 NaN 12.4
1 NaN NaN 13.3
2 11.0 10.8 8.2
Demo dataframe:
import pandas as pd
df = pd.DataFrame({'a': [1,None,3], 'b': [5,10,15]})
I want to replace all NaN values in a with the corresponding values in b**2, and make b NaN (shift NaN values and make some operations on them).
Desired result:
1 5
100 NaN
3 15
How is it possible with pandas?
You can get the rows you want to change using df['a'].isnull(). Then you can use that to update the columns with loc.
import pandas as pd
import numpy as np
df = pd.DataFrame({'a': [1, None, 3], 'b': [5, 10, 15]})
change = df['a'].isnull()
df.loc[change, ['a', 'b']] = [df.loc[change, 'b']**2, np.NaN]
print(df)
Note that the change variable is only to keep from repeating df['a'].isnull() on both sides of the assignment. You could replace it with that expression to do this in one line, but I think that looks cluttered.
Result:
a b
0 1.0 5.0
1 100.0 NaN
2 3.0 15.0
I have the following dataframe (time-series of returns truncated for succinctness):
import pandas as pd
import numpy as np
df = pd.DataFrame({'return':np.array([np.nan, np.nan, np.nan, 0.015, -0.024, 0.033, 0.021, 0.014, -0.092])})
I'm trying to start the index (i.e., "base-100") at the last NaN before the first return - while at the same time keep the NaNs preceding the 100 value in place - (thinking in terms of appending to existing dataframe and for graphing purposes).
I only have found a way to create said index when there are no NaNs in the return vector:
df['index'] = 100*np.exp(np.nan_to_num(df['return'].cumsum()))
Any ideas - thx in advance!
If your initial array is
zz = np.array([np.nan, np.nan, np.nan, 0.015, -0.024, 0.033, 0.021, 0.014, -0.092])
Then you can obtain your desired output like this (although there's probably a more optimized way to do it):
np.concatenate((zz[:np.argmax(np.isfinite(zz))],
100*np.exp(np.cumsum(zz[np.isfinite(zz)]))))
Use Series.isna, change order by indexing and get index of last NaN by Series.idxmax:
idx = df['return'].isna().iloc[::-1].idxmax()
Pass to DataFrame.loc, repalce missing value and use cumulative sum:
df['return'] = df.loc[idx:, 'return'].fillna(100).cumsum()
print (df)
return
0 NaN
1 NaN
2 100.000
3 100.015
4 99.991
5 100.024
6 100.045
7 100.059
8 99.967
You can use Series.isna with Series.cumsum and compare by max, then replace last NaN by Series.fillna and last use cumulative sum:
s = df['return'].isna().cumsum()
df['return'] = df['return'].mask(s.eq(s.max()), df['return'].fillna(100)).cumsum()
print (df)
return
0 NaN
1 NaN
2 100.000
3 100.015
4 99.991
5 100.024
6 100.045
7 100.059
8 99.967
I've got a pandas DataFrame filled mostly with real numbers, but there is a few nan values in it as well.
How can I replace the nans with averages of columns where they are?
This question is very similar to this one: numpy array: replace nan values with average of columns but, unfortunately, the solution given there doesn't work for a pandas DataFrame.
You can simply use DataFrame.fillna to fill the nan's directly:
In [27]: df
Out[27]:
A B C
0 -0.166919 0.979728 -0.632955
1 -0.297953 -0.912674 -1.365463
2 -0.120211 -0.540679 -0.680481
3 NaN -2.027325 1.533582
4 NaN NaN 0.461821
5 -0.788073 NaN NaN
6 -0.916080 -0.612343 NaN
7 -0.887858 1.033826 NaN
8 1.948430 1.025011 -2.982224
9 0.019698 -0.795876 -0.046431
In [28]: df.mean()
Out[28]:
A -0.151121
B -0.231291
C -0.530307
dtype: float64
In [29]: df.fillna(df.mean())
Out[29]:
A B C
0 -0.166919 0.979728 -0.632955
1 -0.297953 -0.912674 -1.365463
2 -0.120211 -0.540679 -0.680481
3 -0.151121 -2.027325 1.533582
4 -0.151121 -0.231291 0.461821
5 -0.788073 -0.231291 -0.530307
6 -0.916080 -0.612343 -0.530307
7 -0.887858 1.033826 -0.530307
8 1.948430 1.025011 -2.982224
9 0.019698 -0.795876 -0.046431
The docstring of fillna says that value should be a scalar or a dict, however, it seems to work with a Series as well. If you want to pass a dict, you could use df.mean().to_dict().
Try:
sub2['income'].fillna((sub2['income'].mean()), inplace=True)
In [16]: df = DataFrame(np.random.randn(10,3))
In [17]: df.iloc[3:5,0] = np.nan
In [18]: df.iloc[4:6,1] = np.nan
In [19]: df.iloc[5:8,2] = np.nan
In [20]: df
Out[20]:
0 1 2
0 1.148272 0.227366 -2.368136
1 -0.820823 1.071471 -0.784713
2 0.157913 0.602857 0.665034
3 NaN -0.985188 -0.324136
4 NaN NaN 0.238512
5 0.769657 NaN NaN
6 0.141951 0.326064 NaN
7 -1.694475 -0.523440 NaN
8 0.352556 -0.551487 -1.639298
9 -2.067324 -0.492617 -1.675794
In [22]: df.mean()
Out[22]:
0 -0.251534
1 -0.040622
2 -0.841219
dtype: float64
Apply per-column the mean of that columns and fill
In [23]: df.apply(lambda x: x.fillna(x.mean()),axis=0)
Out[23]:
0 1 2
0 1.148272 0.227366 -2.368136
1 -0.820823 1.071471 -0.784713
2 0.157913 0.602857 0.665034
3 -0.251534 -0.985188 -0.324136
4 -0.251534 -0.040622 0.238512
5 0.769657 -0.040622 -0.841219
6 0.141951 0.326064 -0.841219
7 -1.694475 -0.523440 -0.841219
8 0.352556 -0.551487 -1.639298
9 -2.067324 -0.492617 -1.675794
Although, the below code does the job, BUT its performance takes a big hit, as you deal with a DataFrame with # records 100k or more:
df.fillna(df.mean())
In my experience, one should replace NaN values (be it with Mean or Median), only where it is required, rather than applying fillna() all over the DataFrame.
I had a DataFrame with 20 variables, and only 4 of them required NaN values treatment (replacement). I tried the above code (Code 1), along with a slightly modified version of it (code 2), where i ran it selectively .i.e. only on variables which had a NaN value
#------------------------------------------------
#----(Code 1) Treatment on overall DataFrame-----
df.fillna(df.mean())
#------------------------------------------------
#----(Code 2) Selective Treatment----------------
for i in df.columns[df.isnull().any(axis=0)]: #---Applying Only on variables with NaN values
df[i].fillna(df[i].mean(),inplace=True)
#---df.isnull().any(axis=0) gives True/False flag (Boolean value series),
#---which when applied on df.columns[], helps identify variables with NaN values
Below is the performance i observed, as i kept on increasing the # records in DataFrame
DataFrame with ~100k records
Code 1: 22.06 Seconds
Code 2: 0.03 Seconds
DataFrame with ~200k records
Code 1: 180.06 Seconds
Code 2: 0.06 Seconds
DataFrame with ~1.6 Million records
Code 1: code kept running endlessly
Code 2: 0.40 Seconds
DataFrame with ~13 Million records
Code 1: --did not even try, after seeing performance on 1.6 Mn records--
Code 2: 3.20 Seconds
Apologies for a long answer ! Hope this helps !
If you want to impute missing values with mean and you want to go column by column, then this will only impute with the mean of that column. This might be a little more readable.
sub2['income'] = sub2['income'].fillna((sub2['income'].mean()))
# To read data from csv file
Dataset = pd.read_csv('Data.csv')
X = Dataset.iloc[:, :-1].values
# To calculate mean use imputer class
from sklearn.impute import SimpleImputer
imputer = SimpleImputer(missing_values=np.nan, strategy='mean')
imputer = imputer.fit(X[:, 1:3])
X[:, 1:3] = imputer.transform(X[:, 1:3])
Directly use df.fillna(df.mean()) to fill all the null value with mean
If you want to fill null value with mean of that column then you can use this
suppose x=df['Item_Weight'] here Item_Weight is column name
here we are assigning (fill null values of x with mean of x into x)
df['Item_Weight'] = df['Item_Weight'].fillna((df['Item_Weight'].mean()))
If you want to fill null value with some string then use
here Outlet_size is column name
df.Outlet_Size = df.Outlet_Size.fillna('Missing')
Pandas: How to replace NaN (nan) values with the average (mean), median or other statistics of one column
Say your DataFrame is df and you have one column called nr_items. This is: df['nr_items']
If you want to replace the NaN values of your column df['nr_items'] with the mean of the column:
Use method .fillna():
mean_value=df['nr_items'].mean()
df['nr_item_ave']=df['nr_items'].fillna(mean_value)
I have created a new df column called nr_item_ave to store the new column with the NaN values replaced by the mean value of the column.
You should be careful when using the mean. If you have outliers is more recommendable to use the median
Another option besides those above is:
df = df.groupby(df.columns, axis = 1).transform(lambda x: x.fillna(x.mean()))
It's less elegant than previous responses for mean, but it could be shorter if you desire to replace nulls by some other column function.
using sklearn library preprocessing class
from sklearn.impute import SimpleImputer
missingvalues = SimpleImputer(missing_values = np.nan, strategy = 'mean', axis = 0)
missingvalues = missingvalues.fit(x[:,1:3])
x[:,1:3] = missingvalues.transform(x[:,1:3])
Note: In the recent version parameter missing_values value change to np.nan from NaN
I use this method to fill missing values by average of a column.
fill_mean = lambda col : col.fillna(col.mean())
df = df.apply(fill_mean, axis = 0)
You can also use value_counts to get the most frequent values. This would work on different datatypes.
df = df.apply(lambda x:x.fillna(x.value_counts().index[0]))
Here is the value_counts api reference.