CalZone
0
Q:

pandas compare two columns of different dataframe

comparison_column = np.where(df["col1"] == df["col2"], True, False)
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#suppose you have two dataframes df1 and df2, and 
#you need to merge them along the column id
df_merge_col = pd.merge(df1, df2, on='id')
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# 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.



1
df1['enh2'] = pd.Series((df2.type.isin(df1.type)) & (df1.value != df2.low)  | (df1.value + 1 == df2.high))
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import numpy as np
df1['low_value'] = np.where(df1.value <= df2.low, 'True', 'False')
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