Pandas "Can only compare identically-labeled DataFrame objects" error

I'm using Pandas to compare the outputs of two files loaded into two data frames (uat, prod): ...

uat = uat[['Customer Number','Product']] prod = prod[['Customer Number','Product']] print uat['Customer Number'] == prod['Customer Number'] print uat['Product'] == prod['Product'] print uat == prod The first two match exactly: 74357 True 74356 True Name: Customer Number, dtype: bool 74357 True 74356 True Name: Product, dtype: bool 

For the third print, I get an error: Can only compare identically-labeled DataFrame objects. If the first two compared fine, what's wrong with the 3rd?

Thanks

7 Answers

Here's a small example to demonstrate this (which only applied to DataFrames, not Series, until Pandas 0.19 where it applies to both):

In [1]: df1 = pd.DataFrame([[1, 2], [3, 4]]) In [2]: df2 = pd.DataFrame([[3, 4], [1, 2]], index=[1, 0]) In [3]: df1 == df2 Exception: Can only compare identically-labeled DataFrame objects 

One solution is to sort the index first (Note: some functions require sorted indexes):

In [4]: df2.sort_index(inplace=True) In [5]: df1 == df2 Out[5]: 0 1 0 True True 1 True True 

Note: == is also sensitive to the order of columns, so you may have to use sort_index(axis=1):

In [11]: df1.sort_index().sort_index(axis=1) == df2.sort_index().sort_index(axis=1) Out[11]: 0 1 0 True True 1 True True 

Note: This can still raise (if the index/columns aren't identically labelled after sorting).

You can also try dropping the index column if it is not needed to compare:

print(df1.reset_index(drop=True) == df2.reset_index(drop=True)) 

I have used this same technique in a unit test like so:

from pandas.util.testing import assert_frame_equal assert_frame_equal(actual.reset_index(drop=True), expected.reset_index(drop=True)) 

At the time when this question was asked there wasn't another function in Pandas to test equality, but it has been added a while ago: pandas.equals

You use it like this:

df1.equals(df2) 

Some differenes to == are:

• You don't get the error described in the question
• It returns a simple boolean.
• NaN values in the same location are considered equal
• 2 DataFrames need to have the same dtype to be considered equal, see this stackoverflow question

EDIT:
As pointed out in @paperskilltrees answer index alignment is important. Apart from the solution provided there another option is to sort the index of the DataFrames before comparing the DataFrames. For df1 that would be df1.sort_index(inplace=True).

When you compare two DataFrames, you must ensure that the number of records in the first DataFrame matches with the number of records in the second DataFrame. In our example, each of the two DataFrames had 4 records, with 4 products and 4 prices.

If, for example, one of the DataFrames had 5 products, while the other DataFrame had 4 products, and you tried to run the comparison, you would get the following error:

ValueError: Can only compare identically-labeled Series objects

this should work

import pandas as pd import numpy as np firstProductSet = {'Product1': ['Computer','Phone','Printer','Desk'], 'Price1': [1200,800,200,350] } df1 = pd.DataFrame(firstProductSet,columns= ['Product1', 'Price1']) secondProductSet = {'Product2': ['Computer','Phone','Printer','Desk'], 'Price2': [900,800,300,350] } df2 = pd.DataFrame(secondProductSet,columns= ['Product2', 'Price2']) df1['Price2'] = df2['Price2'] #add the Price2 column from df2 to df1 df1['pricesMatch?'] = np.where(df1['Price1'] == df2['Price2'], 'True', 'False') #create new column in df1 to check if prices match df1['priceDiff?'] = np.where(df1['Price1'] == df2['Price2'], 0, df1['Price1'] - df2['Price2']) #create new column in df1 for price diff print (df1) 

example from

Flyingdutchman's answer is great but wrong: it uses DataFrame.equals, which will return False in your case. Instead, you want to use DataFrame.eq, which will return True.

It seems that DataFrame.equals ignores the dataframe's index, while DataFrame.eq uses dataframes' indexes for alignment and then compares the aligned values. This is an occasion to quote the central gotcha of Pandas:

Here is a basic tenet to keep in mind: data alignment is intrinsic. The link between labels and data will not be broken unless done so explicitly by you.

As we can see in the following examples, the data alignment is neither broken, nor enforced, unless explicitly requested. So we have three different situations.

1. No explicit instruction given, as to the alignment: == aka DataFrame.__eq__,

 In [1]: import pandas as pd In [2]: df1 = pd.DataFrame(index=[0, 1, 2], data={'col1':list('abc')}) In [3]: df2 = pd.DataFrame(index=[2, 0, 1], data={'col1':list('cab')}) In [4]: df1 == df2 --------------------------------------------------------------------------- ... ValueError: Can only compare identically-labeled DataFrame objects 

2. Alignment is explicitly broken: DataFrame.equals, DataFrame.values, DataFrame.reset_index(),

 In [5]: df1.equals(df2) Out[5]: False In [9]: df1.values == df2.values Out[9]: array([[False], [False], [False]]) In [10]: (df1.values == df2.values).all().all() Out[10]: False 

3. Alignment is explicitly enforced: DataFrame.eq, DataFrame.sort_index(),

 In [6]: df1.eq(df2) Out[6]: col1 0 True 1 True 2 True In [8]: df1.eq(df2).all().all() Out[8]: True 

My answer is as of pandas version 1.0.3.

Here I am showing a complete example of how to handle this error. I have added rows with zeros. You can have your dataframes from csv or any other source.

import pandas as pd import numpy as np # df1 with 9 rows df1 = pd.DataFrame({'Name':['John','Mike','Smith','Wale','Marry','Tom','Menda','Bolt','Yuswa',], 'Age':[23,45,12,34,27,44,28,39,40]}) # df2 with 8 rows df2 = pd.DataFrame({'Name':['John','Mike','Wale','Marry','Tom','Menda','Bolt','Yuswa',], 'Age':[25,45,14,34,26,44,29,42]}) # get lengths of df1 and df2 df1_len = len(df1) df2_len = len(df2) diff = df1_len - df2_len rows_to_be_added1 = rows_to_be_added2 = 0 # rows_to_be_added1 = np.zeros(diff) if diff < 0: rows_to_be_added1 = abs(diff) else: rows_to_be_added2 = diff # add empty rows to df1 if rows_to_be_added1 > 0: df1 = df1.append(pd.DataFrame(np.zeros((rows_to_be_added1,len(df1.columns))),columns=df1.columns)) # add empty rows to df2 if rows_to_be_added2 > 0: df2 = df2.append(pd.DataFrame(np.zeros((rows_to_be_added2,len(df2.columns))),columns=df2.columns)) # at this point we have two dataframes with the same number of rows, and maybe different indexes # drop the indexes of both, so we can compare the dataframes and other operations like update etc. df2.reset_index(drop=True, inplace=True) df1.reset_index(drop=True, inplace=True) # add a new column to df1 df1['New_age'] = None # compare the Age column of df1 and df2, and update the New_age column of df1 with the Age column of df2 if they match, else None df1['New_age'] = np.where(df1['Age'] == df2['Age'], df2['Age'], None) # drop rows where Name is 0.0 df2 = df2.drop(df2[df2['Name'] == 0.0].index) # now we don't get the error ValueError: Can only compare identically-labeled Series objects 

I found where the error is coming from in my case:

The problem was that column names list was accidentally enclosed in another list.

Consider following example:

column_names=['warrior','eat','ok','monkeys'] df_good = pd.DataFrame(np.ones(shape=(6,4)),columns=column_names) df_good['ok'] < df_good['monkeys'] >>> 0 False 1 False 2 False 3 False 4 False 5 False df_bad = pd.DataFrame(np.ones(shape=(6,4)),columns=[column_names]) df_bad ['ok'] < df_bad ['monkeys'] >>> ValueError: Can only compare identically-labeled DataFrame objects 

And the thing is you cannot visually distinguish the bad DataFrame from good.