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.
I have a pandas series as follows...
0 2039-03-16
1 2056-01-21
2 2051-11-18
3 2064-03-05
4 2048-06-05
Name: BIRTH, dtype: datetime64
It was created from string data as follows
s = data['BIRTH']
s = pd.to_datetime(s)
s
I want to convert all dates after year 2040 to 1940
I can do this for a single record as follows
s.iloc[0].replace(year=d.year-100)
but I really want to just run it over the whole series. I can't work it out. Help!??
PS - I know there's ways outside of pandas using Python's DT module but I'd like to learn how to do this within Pandas please
Using DateOffset is the obvious choice here:
df['date'] - pd.offsets.DateOffset(years=100)
0 1939-03-16
1 1956-01-21
2 1951-11-18
3 1964-03-05
4 1948-06-05
Name: date, dtype: datetime64[ns]
Assign it back:
df['date'] -= pd.offsets.DateOffset(years=100)
df
date
0 1939-03-16
1 1956-01-21
2 1951-11-18
3 1964-03-05
4 1948-06-05
We have the offsets module to deal with non-fixed frequencies, it comes in handy in situations like these.
To fix your code, you'd have wanted to apply datetime.replace rowwise using apply (not recommended):
df['date'].apply(lambda x: x.replace(year=x.year-100))
0 1939-03-16
1 1956-01-21
2 1951-11-18
3 1964-03-05
4 1948-06-05
Name: date, dtype: datetime64[ns]
Or using a list comprehension,
df.assign(date=[x.replace(year=x.year-100) for x in df['date']])
date
0 1939-03-16
1 1956-01-21
2 1951-11-18
3 1964-03-05
4 1948-06-05
Neither of these handle NaT entries very well.
I have got a datetime variable in pandas dataframe 1, when I check the dtypes, it shows the right format (datetime) [2], however when I try to plot this variable, it is being plotted as numbers and not datetime [3].
The most surprising is that this variable was working fine till yesterday, I do not know what has change today :( and as the dtype is showing fine, I am clueless what else could go wrong.
I would highly appreciate your feedback.
thank you,
1
df.head()
reactive_power current timeofmeasurement
0 0 0.000 2018-12-12 10:43:41
1 0 0.000 2018-12-12 10:44:32
2 0 1.147 2018-12-12 10:46:16
3 262 1.135 2018-12-12 10:47:30
4 1159 4.989 2018-12-12 10:49:47
[2]
[] df.dtypes
reactive_power int64
current float64
timeofmeasurement datetime64[ns]
dtype: object
[3]
[]1
You need to convert your datetime column from string type into datetime type, and then set it as index. I don't have your original code, but something along those lines:
#Convert to datetime
df["current timeofmeasurement"] = pd.to_datetime(df["current timeofmeasurement"], format = "%Y-%m-%d %H:%H:%S")
#Set date as index
df = df.set_index("current timeofmeasurement")
#Then you can plot easily
df.plot()
I have df with index as date and also column called scores. Now I want to maintain the df as it is but add column which gives the 0.7 quantile of scores for that day. Method of quantile would need to be midpoint and also be rounded to nearest whole number.
I've outlined one approach you could take, below.
Note that to round a value to the nearest whole number you should use Python's built-in round() function. See round() in the Python documentation for details.
import pandas as pd
import numpy as np
# set random seed for reproducibility
np.random.seed(748)
# initialize base example dataframe
df = pd.DataFrame({"date":np.arange(10),
"score":np.random.uniform(size=10)})
duplicate_dates = np.random.choice(df.index, 5)
df_dup = pd.DataFrame({"date":np.random.choice(df.index, 5),
"score":np.random.uniform(size=5)})
# finish compiling example data
df = df.append(df_dup, ignore_index=True)
# calculate 0.7 quantile result with specified parameters
result = df.groupby("date").quantile(q=0.7, axis=0, interpolation='midpoint')
# print resulting dataframe
# contains one unique 0.7 quantile value per date
print(result)
"""
0.7 score
date
0 0.585087
1 0.476404
2 0.426252
3 0.363376
4 0.165013
5 0.927199
6 0.575510
7 0.576636
8 0.831572
9 0.932183
"""
# to apply the resulting quantile information to
# a new column in our original dataframe `df`
# we can apply a dictionary to our "date" column
# create dictionary
mapping = result.to_dict()["score"]
# apply to `df` to produce desired new column
df["quantile_0.7"] = [mapping[x] for x in df["date"]]
print(df)
"""
date score quantile_0.7
0 0 0.920895 0.585087
1 1 0.476404 0.476404
2 2 0.380771 0.426252
3 3 0.363376 0.363376
4 4 0.165013 0.165013
5 5 0.927199 0.927199
6 6 0.340008 0.575510
7 7 0.695818 0.576636
8 8 0.831572 0.831572
9 9 0.932183 0.932183
10 7 0.457455 0.576636
11 6 0.650666 0.575510
12 6 0.500353 0.575510
13 0 0.249280 0.585087
14 2 0.471733 0.426252
"""
I have a numpy series with values like "1.0s", "100ms", etc. I can't plot this (with pandas, after putting the array into a series), because pandas doesn't recognize that these are numbers. How can I have numpy or pandas extrapolate these into numbers, while paying attention to the suffixes?
see question how do I get at the pandas.offsets object given an offset string
use pandas.tseries.frequencies.to_offset
convert to timedeltas
get total seconds
from pandas.tseries.frequencies import to_offset
s = pd.Series(['1.0s', '100ms', '10s', '0.5T'])
pd.to_timedelta(s.apply(to_offset)).dt.total_seconds()
0 0.0
1 0.1
2 10.0
3 300.0
dtype: float64
This code could solve your problem.
# Test data
se = Series(['10s', '100ms', '1.0s'])
# Pattern to match ms and as integer of float
pat = "([0-9]*\.?[0-9]+)(ms|s)"
# Extracting the data
df = se.str.extract(pat, flags=0, expand=True)
# Renaming columns
df.columns = ['value', 'unit']
# Converting to number
df['value'] = pd.to_numeric(df['value'])
# Converting to the same unit
df.loc[df['unit']=='s', ['value', 'unit']] = (df['value'] * 1000, 'ms')
# Now you are ready to plot !
print(df['value'])
# 0 10000.0
# 1 100.0
# 2 100000.0