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.
Related
Both tables that I merge have the cells formatted correctly, as numbers, but when I make a left join, the numbers in one of the original tables get dis-formatted (you see e+ in those numbers). What should I do to see those numbers un full?
Problem: When merging, some SKU values that appear in df1 do not appear in df2. In order to represent unavailable values, pandas automatically uses NaN, which is a floating point value. Thus, the integer ISBNs are converted to float. Given the size of the ISBNs, pandas then formats these floating point values in scientific notation.
You could solve this by defining your own floating point value formatter (pd.options.display.float_format), but in your case it might be easier / more effective to convert the ISBNs to a string before merging.
Example:
>>> import pandas as pd
>>> df1 = pd.DataFrame({"SKU": list("abcde"), "ISBN": list(range(1, 6))})
>>> df2 = pd.DataFrame({"SKU": list("bcef"), "ISBN": list(range(4, 8))})
Your problem:
>>> pd.merge(df1, df2, on="SKU", how="left")
SKU ISBN_x ISBN_y
0 a 1 NaN
1 b 2 4.0
2 c 3 5.0
3 d 4 NaN
4 e 5 6.0
>>> _.dtypes
SKU object
ISBN_x int64
ISBN_y float64 # <<< Problematic
vs possible solution:
>>> pd.merge(df1.astype(str), df2.astype(str), on="SKU", how="left")
SKU ISBN_x ISBN_y
0 a 1 NaN
1 b 2 4
2 c 3 5
3 d 4 NaN
4 e 5 6
>>> _.dtypes
SKU object
ISBN_x object
ISBN_y object
I have a situation where I need to merge multiple dataframes that I can do easily using the below code:
# Merge all the datasets together
df_prep1 = df_prep.merge(df1,on='e_id',how='left')
df_prep2 = df_prep1.merge(df2,on='e_id',how='left')
df_prep3 = df_prep2.merge(df3,on='e_id',how='left')
df_prep4 = df_prep3.merge(df_4,on='e_id',how='left')
df_prep5 = df_prep4.merge(df_5,on='e_id',how='left')
df_prep6 = df_prep5.merge(df_6,on='e_id',how='left')
But what I want to understand is that if there is any other efficient way to perform this merge, maybe using a helper function? If yes, then how could I achieve that?
You can use reduce from functools module to merge multiple dataframes:
from functools import reduce
dfs = [df_1, df_2, df_3, df_4, df_5, df_6]
out = reduce(lambda dfl, dfr: pd.merge(dfl, dfr, on='e_id', how='left'), dfs)
You can put all your dfs into a list, or pass them from a function, a loop, etc. and then have 1 main df that you merge everything onto.
You can start with an empty df and iterate through. In your case, since you are doing left merge, it looks like your df_prep should already have all of the e_id values that you want. You'll need to figure out what you want to do with any additional columns, e.g., you can have pandas add _x and _y after conflicting column names that you don't merge, or rename them, etc. See this toy example:
main_df = pd.DataFrame({'e_id': [0, 1, 2, 3, 4]})
for x in range(3):
dfx = pd.DataFrame({'e_id': [x], 'another_col' + str(x): [x * 10]})
main_df = main_df.merge(dfx, on='e_id', how='left')
to get:
e_id another_col0 another_col1 another_col2
0 0 0.0 NaN NaN
1 1 NaN 10.0 NaN
2 2 NaN NaN 20.0
3 3 NaN NaN NaN
4 4 NaN NaN NaN
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
Why do simple DataFrame op DataFrame operations result in a union'ed DataFrame? Pandas documentation mentions unionizing because of alignment issues. I don't see any alignment issues with df1 and df2. Aren't alignment issues about different shapes, different dtypes, or different indexes?
df1 = pd.DataFrame([[1,2],[3,4]],columns=list('AB'))
df2 = pd.DataFrame([[5,6],[7,8]],columns=list('CD'))
>> df1*df2
A B C D
0 NaN NaN NaN NaN
1 NaN NaN NaN NaN
Another source of alignment issues is non-matching column names. Here, alignment requires identical column names. Either make the column names the same or use .values. Using .values on just the right-hand DataFrame will retain the DataFrame type.
>> df1*df2.values
A B
0 5 12
1 21 32
I have a pandas dataframe that has columns:
'video' and 'link' of click values
with an index of datetime. For some reason, when I use semilogy and boxplot with the video series, I get the error
ValueError: Data has no positive values, and therefore can not be log-scaled.
but when I do it on the 'link' series I can draw the boxplot correctly.
I have verified that both the 'video' and 'link' series has NaN values and positive values.
Any thoughts on why this is occurring? Below is what I've done to verify that this is the case
Below is sample code:
#get all the not null values of video to show that there are positive
temp=a.types_pivot[a.types_pivot['video'].notnull()]
print temp
#get a count of all the NaN values to show both 'video' and 'link' has NaN
count = 0
for item in a.types_pivot['video']:
if(item.is_integer() == False):
count += 1
#try to draw the plots
print "there is %s nan values in video" % (count)
fig=plt.figure(figsize=(6,6),dpi=50)
ax=fig.add_subplot(111)
ax.semilogy()
plt.boxplot(a.types_pivot['video'].values)
Here is relevant output from the code for video series
type link video
created_time
2011-02-10 15:00:51+00:00 NaN 5
2011-02-17 17:50:38+00:00 NaN 5
2011-03-22 14:04:56+00:00 NaN 5
there is 5463 nan values in video
I run the same exact code except I do
a.types_pivot['link']
and I am able to draw the boxplot.
Below is the relevant output from the link series
Index: 5269 entries, 2011-01-24 20:03:58+00:00 to 2012-06-22 16:56:30+00:00
Data columns:
link 5269 non-null values
photo 0 non-null values
question 0 non-null values
status 0 non-null values
swf 0 non-null values
video 0 non-null values
dtypes: float64(6)
there is 216 nan values in link
Using the describe function
a.types_pivot['video'].describe()
<pre>
count 22.000000
mean 16.227273
std 15.275040
min 1.000000
25% 5.250000
50% 9.500000
75% 23.000000
max 58.000000
</pre>
Note: I'm unable to upload images due to some issue with imgur. I'll try again later.
Take advantage of pandas matplotlib helper / wrappers by calling pd.DataFrame.boxplot(). I believe this will take care of the NaN values for you. It will also put both Series in the same plot so you can easily compare data.
Example
Create a dataframe with some NaN values and negative values
In [7]: df = pd.DataFrame(np.random.rand(10, 5))
In [8]: df.ix[2:4,3] = np.nan
In [9]: df.ix[2:3,4] = -0.45
In [10]: df
Out[10]:
0 1 2 3 4
0 0.391882 0.776331 0.875009 0.350585 0.154517
1 0.772635 0.657556 0.745614 0.725191 0.483967
2 0.057269 0.417439 0.861274 NaN -0.450000
3 0.997749 0.736229 0.084077 NaN -0.450000
4 0.886303 0.596473 0.943397 NaN 0.816650
5 0.018724 0.459743 0.472822 0.598056 0.273341
6 0.894243 0.097513 0.691781 0.802758 0.785258
7 0.222901 0.292646 0.558909 0.220400 0.622068
8 0.458428 0.039280 0.670378 0.457238 0.912308
9 0.516554 0.445004 0.356060 0.861035 0.433503
Note that I can count the number of NaN values like so:
In [14]: df[3].isnull().sum() # Count NaNs in the 4th column
Out[14]: 3
A box plot is simply:
In [16]: df.boxplot()
You could create a semi-log boxplot, for example, by:
In [23]: np.log(df).boxplot()
Or, more generally, modify / transform to you heart's content, and then boxplot.
In [24]: df_mod = np.log(df).dropna()
In [25]: df_mod.boxplot()