pandas compare dataframes code example

Example 1: python pandas difference between two data frames

diff_df = pd.merge(df1, df2, how='outer', indicator='Exist')

diff_df = diff_df.loc[diff_df['Exist'] != 'both']

Example 2: comparing two dataframe columns

comparison_column = np.where(df["col1"] == df["col2"], True, False)

Example 3: dataframe equals pandas

>>> exactly_equal = pd.DataFrame({1: [10], 2: [20]})
>>> exactly_equal
    1   2
0  10  20
>>> df.equals(exactly_equal)
True

Example 4: pandas compare two columns

# Syntax:
# C = np.where(condition, A, B) 
# equal to A when condition true and B when false
import numpy as np
import pandas as pd

a = [['10', '1.2', '4.2'], ['15', '70', '0.03'], ['8', '5', '0']]
df = pd.DataFrame(a, columns=['one', 'two', 'three'])

df['que'] = np.where((df['one'] >= df['two']) & (df['one'] <= df['three'])
                     , df['one'], np.nan)

# If you have more than one condition, then you could use np.select 
# instead. For example, if you wish df['que'] to equal 
# df['two'] when df['one'] < df['two'], then
conditions = [
    (df['one'] >= df['two']) & (df['one'] <= df['three']), 
    df['one'] < df['two']]

choices = [df['one'], df['two']]

df['que'] = np.select(conditions, choices, default=np.nan)

# If we can assume that df['one'] >= df['two'] when 
# df['one'] < df['two'] is False, then the conditions and 
# choices could be simplified to
conditions = [
    df['one'] < df['two'],
    df['one'] <= df['three']]

choices = [df['two'], df['one']]

# Note that: 
a = [['10', '1.2', '4.2'], ['15', '70', '0.03'], ['8', '5', '0']]
df = pd.DataFrame(a, columns=['one', 'two', 'three'])

# defines a DataFrame with string values. Since they look numeric, 
# you might be better off converting those strings to floats:
df2 = df.astype(float)

# This changes the results, however, since strings compare 
# character-by-character, while floats are compared numerically.

Example 5: pandas two dataframes equal

>>> df = pd.DataFrame({1: [10], 2: [20]})

>>> exactly_equal = pd.DataFrame({1: [10], 2: [20]}) # Note the same as above

>>> df.equals(exactly_equal)
True
# Two are the same hence true. If not would be false