Compare multiple columns in two dataframes and select rows with differing values
0
I'm trying to compare 2 columns in one dataframe(df1) with 2 columns in another dataframe(df2). After comparison, I want to select the rows where the first two columns do not match. You can see my attempts below and this what the dataframes look like [1]
import pandas as pd
fd1= 'Q37.xlsx'
fd2= 'Q43.xlsx'
df1 = pd.read_excel( fd1, sheetname='prio 1')
df2 = pd.read_excel( fd2, sheetname='prio 1')
closed_items= {} #items in fd1 but not in fd2
new_items={} #items in fd2 but not in fd1
In order to get closed_items, I've tried the following 3 things
closed_items.where(df1[df1['Code'].values!=df2[df2['Code'].values and
df1['Owner'].values != key in df1['Owner'].values)
and gotten
ValueError: Can only compare identically-labeled Series objects
I've also tried
Closed_items = df2.loc[(df2['Code'] != df1['Code']) and
df2.loc[(df2['Owner'] != df1['Owner'])]
And lastly I tried
for key in df1['Code'].values:
if key in df1['Code'].values != key in df1['Code'].values or key in
df1['Owner'].values != key in df1['Owner'].values:
closed_items.append()
else:
pass
Which gave this syntax
The truth value of an array with more than one element is ambiguous.
Use a.any() or a.all()
...
AFP= pd.ExcelWriter("AFP.xlsx", engine='xlsxwriter')
closed_items.to_excel(AFP, sheet_name='Closed', index=False)
python excel pandas dataframe conditional-formatting
edited Nov 27 '18 at 13:07
user3471881
1,1922619
asked Nov 27 '18 at 12:02
gabriella gabriella
185
add a comment |
0
I'm trying to compare 2 columns in one dataframe(df1) with 2 columns in another dataframe(df2). After comparison, I want to select the rows where the first two columns do not match. You can see my attempts below and this what the dataframes look like [1]
import pandas as pd
fd1= 'Q37.xlsx'
fd2= 'Q43.xlsx'
df1 = pd.read_excel( fd1, sheetname='prio 1')
df2 = pd.read_excel( fd2, sheetname='prio 1')
closed_items= {} #items in fd1 but not in fd2
new_items={} #items in fd2 but not in fd1
In order to get closed_items, I've tried the following 3 things
closed_items.where(df1[df1['Code'].values!=df2[df2['Code'].values and
df1['Owner'].values != key in df1['Owner'].values)
and gotten
ValueError: Can only compare identically-labeled Series objects
I've also tried
Closed_items = df2.loc[(df2['Code'] != df1['Code']) and
df2.loc[(df2['Owner'] != df1['Owner'])]
And lastly I tried
for key in df1['Code'].values:
if key in df1['Code'].values != key in df1['Code'].values or key in
df1['Owner'].values != key in df1['Owner'].values:
closed_items.append()
else:
pass
Which gave this syntax
The truth value of an array with more than one element is ambiguous.
Use a.any() or a.all()
...
AFP= pd.ExcelWriter("AFP.xlsx", engine='xlsxwriter')
closed_items.to_excel(AFP, sheet_name='Closed', index=False)
python excel pandas dataframe conditional-formatting
edited Nov 27 '18 at 13:07
user3471881
1,1922619
asked Nov 27 '18 at 12:02
gabriella gabriella
185
I would think better is to have equally shaped series and try to use something like not s1.intersection(s2) or difference stackoverflow.com/questions/18079563/… pandas.pydata.org/pandas-docs/stable/generated/…
– Sergii
Nov 27 '18 at 12:18
add a comment |
0
0
0
1
I'm trying to compare 2 columns in one dataframe(df1) with 2 columns in another dataframe(df2). After comparison, I want to select the rows where the first two columns do not match. You can see my attempts below and this what the dataframes look like [1]
import pandas as pd
fd1= 'Q37.xlsx'
fd2= 'Q43.xlsx'
df1 = pd.read_excel( fd1, sheetname='prio 1')
df2 = pd.read_excel( fd2, sheetname='prio 1')
closed_items= {} #items in fd1 but not in fd2
new_items={} #items in fd2 but not in fd1
In order to get closed_items, I've tried the following 3 things
closed_items.where(df1[df1['Code'].values!=df2[df2['Code'].values and
df1['Owner'].values != key in df1['Owner'].values)
and gotten
ValueError: Can only compare identically-labeled Series objects
I've also tried
Closed_items = df2.loc[(df2['Code'] != df1['Code']) and
df2.loc[(df2['Owner'] != df1['Owner'])]
And lastly I tried
for key in df1['Code'].values:
if key in df1['Code'].values != key in df1['Code'].values or key in
df1['Owner'].values != key in df1['Owner'].values:
closed_items.append()
else:
pass
Which gave this syntax
The truth value of an array with more than one element is ambiguous.
Use a.any() or a.all()
...
AFP= pd.ExcelWriter("AFP.xlsx", engine='xlsxwriter')
closed_items.to_excel(AFP, sheet_name='Closed', index=False)
python excel pandas dataframe conditional-formatting
edited Nov 27 '18 at 13:07
user3471881
1,1922619
asked Nov 27 '18 at 12:02
gabriella gabriella
185
I'm trying to compare 2 columns in one dataframe(df1) with 2 columns in another dataframe(df2). After comparison, I want to select the rows where the first two columns do not match. You can see my attempts below and this what the dataframes look like [1]
import pandas as pd
fd1= 'Q37.xlsx'
fd2= 'Q43.xlsx'
df1 = pd.read_excel( fd1, sheetname='prio 1')
df2 = pd.read_excel( fd2, sheetname='prio 1')
closed_items= {} #items in fd1 but not in fd2
new_items={} #items in fd2 but not in fd1
In order to get closed_items, I've tried the following 3 things
closed_items.where(df1[df1['Code'].values!=df2[df2['Code'].values and
df1['Owner'].values != key in df1['Owner'].values)
and gotten
ValueError: Can only compare identically-labeled Series objects
I've also tried
Closed_items = df2.loc[(df2['Code'] != df1['Code']) and
df2.loc[(df2['Owner'] != df1['Owner'])]
And lastly I tried
for key in df1['Code'].values:
if key in df1['Code'].values != key in df1['Code'].values or key in
df1['Owner'].values != key in df1['Owner'].values:
closed_items.append()
else:
pass
Which gave this syntax
The truth value of an array with more than one element is ambiguous.
Use a.any() or a.all()
...
AFP= pd.ExcelWriter("AFP.xlsx", engine='xlsxwriter')
closed_items.to_excel(AFP, sheet_name='Closed', index=False)
python excel pandas dataframe conditional-formatting
python excel pandas dataframe conditional-formatting
edited Nov 27 '18 at 13:07
user3471881
1,1922619
asked Nov 27 '18 at 12:02
gabriella gabriella
185
edited Nov 27 '18 at 13:07
user3471881
1,1922619
asked Nov 27 '18 at 12:02
gabriella gabriella
185
edited Nov 27 '18 at 13:07
user3471881
1,1922619
edited Nov 27 '18 at 13:07
user3471881
1,1922619
edited Nov 27 '18 at 13:07
user3471881
1,1922619
1,1922619
asked Nov 27 '18 at 12:02
gabriella gabriella
185
asked Nov 27 '18 at 12:02
gabriella gabriella
185
asked Nov 27 '18 at 12:02
gabriella gabriella
185
185
I would think better is to have equally shaped series and try to use something like not s1.intersection(s2) or difference stackoverflow.com/questions/18079563/… pandas.pydata.org/pandas-docs/stable/generated/…
– Sergii
Nov 27 '18 at 12:18
add a comment |
I would think better is to have equally shaped series and try to use something like not s1.intersection(s2) or difference stackoverflow.com/questions/18079563/… pandas.pydata.org/pandas-docs/stable/generated/…
– Sergii
Nov 27 '18 at 12:18
I would think better is to have equally shaped series and try to use something like not s1.intersection(s2) or difference stackoverflow.com/questions/18079563/… pandas.pydata.org/pandas-docs/stable/generated/…
– Sergii
Nov 27 '18 at 12:18
I would think better is to have equally shaped series and try to use something like not s1.intersection(s2) or difference stackoverflow.com/questions/18079563/… pandas.pydata.org/pandas-docs/stable/generated/…
– Sergii
Nov 27 '18 at 12:18
add a comment |
1 Answer
1
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votes
0
The problem is that df1 and df2 are of different shapes hence the loc will not work.
You first need to merge df1 and df2 like
df3 = df1.merge(df2,on='common_key',how='left',suffixes=('_df1','_df2'))
df3['select'] = 0
df3.loc[(df3['Code_df1'] == df3['Code_df2']) &
(df3.loc[(df3['Owner_df1'] == df3['Owner_df2']),'select'])] = 1
df3.loc[df3['select']==0,:]
will return wherever they do not match
answered Nov 27 '18 at 12:11
Abhishek SharmaAbhishek Sharma
6371622
It returns Invalid syntax on df3.loc[(df3['Code_df1'] == df3['Code_df2']) & df3.loc[(df3['Owner_df1'] == df3['Owner_df2']),'select'] = 1
– gabriella
Nov 29 '18 at 8:50
Use this I missed a closing parenthesis while copying the code. df3.loc[(df3['Code_df1'] == df3['Code_df2']) & (df3.loc[(df3['Owner_df1'] == df3['Owner_df2']),'select'])] = 1
– Abhishek Sharma
Nov 30 '18 at 9:10
add a comment |
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1 Answer
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The problem is that df1 and df2 are of different shapes hence the loc will not work.
You first need to merge df1 and df2 like
df3 = df1.merge(df2,on='common_key',how='left',suffixes=('_df1','_df2'))
df3['select'] = 0
df3.loc[(df3['Code_df1'] == df3['Code_df2']) &
(df3.loc[(df3['Owner_df1'] == df3['Owner_df2']),'select'])] = 1
df3.loc[df3['select']==0,:]
will return wherever they do not match
answered Nov 27 '18 at 12:11
Abhishek SharmaAbhishek Sharma
6371622
It returns Invalid syntax on df3.loc[(df3['Code_df1'] == df3['Code_df2']) & df3.loc[(df3['Owner_df1'] == df3['Owner_df2']),'select'] = 1
– gabriella
Nov 29 '18 at 8:50
Use this I missed a closing parenthesis while copying the code. df3.loc[(df3['Code_df1'] == df3['Code_df2']) & (df3.loc[(df3['Owner_df1'] == df3['Owner_df2']),'select'])] = 1
– Abhishek Sharma
Nov 30 '18 at 9:10
add a comment |
0
The problem is that df1 and df2 are of different shapes hence the loc will not work.
You first need to merge df1 and df2 like
df3 = df1.merge(df2,on='common_key',how='left',suffixes=('_df1','_df2'))
df3['select'] = 0
df3.loc[(df3['Code_df1'] == df3['Code_df2']) &
(df3.loc[(df3['Owner_df1'] == df3['Owner_df2']),'select'])] = 1
df3.loc[df3['select']==0,:]
will return wherever they do not match
answered Nov 27 '18 at 12:11
Abhishek SharmaAbhishek Sharma
6371622
It returns Invalid syntax on df3.loc[(df3['Code_df1'] == df3['Code_df2']) & df3.loc[(df3['Owner_df1'] == df3['Owner_df2']),'select'] = 1
– gabriella
Nov 29 '18 at 8:50
Use this I missed a closing parenthesis while copying the code. df3.loc[(df3['Code_df1'] == df3['Code_df2']) & (df3.loc[(df3['Owner_df1'] == df3['Owner_df2']),'select'])] = 1
– Abhishek Sharma
Nov 30 '18 at 9:10
add a comment |
0
0
0
The problem is that df1 and df2 are of different shapes hence the loc will not work.
You first need to merge df1 and df2 like
df3 = df1.merge(df2,on='common_key',how='left',suffixes=('_df1','_df2'))
df3['select'] = 0
df3.loc[(df3['Code_df1'] == df3['Code_df2']) &
(df3.loc[(df3['Owner_df1'] == df3['Owner_df2']),'select'])] = 1
df3.loc[df3['select']==0,:]
will return wherever they do not match
answered Nov 27 '18 at 12:11
Abhishek SharmaAbhishek Sharma
6371622
The problem is that df1 and df2 are of different shapes hence the loc will not work.
You first need to merge df1 and df2 like
df3 = df1.merge(df2,on='common_key',how='left',suffixes=('_df1','_df2'))
df3['select'] = 0
df3.loc[(df3['Code_df1'] == df3['Code_df2']) &
(df3.loc[(df3['Owner_df1'] == df3['Owner_df2']),'select'])] = 1
df3.loc[df3['select']==0,:]
will return wherever they do not match
answered Nov 27 '18 at 12:11
Abhishek SharmaAbhishek Sharma
6371622
edited Nov 30 '18 at 9:13
answered Nov 27 '18 at 12:11
Abhishek SharmaAbhishek Sharma
6371622
answered Nov 27 '18 at 12:11
Abhishek SharmaAbhishek Sharma
6371622
answered Nov 27 '18 at 12:11
Abhishek SharmaAbhishek Sharma
6371622
6371622
It returns Invalid syntax on df3.loc[(df3['Code_df1'] == df3['Code_df2']) & df3.loc[(df3['Owner_df1'] == df3['Owner_df2']),'select'] = 1
– gabriella
Nov 29 '18 at 8:50
Use this I missed a closing parenthesis while copying the code. df3.loc[(df3['Code_df1'] == df3['Code_df2']) & (df3.loc[(df3['Owner_df1'] == df3['Owner_df2']),'select'])] = 1
– Abhishek Sharma
Nov 30 '18 at 9:10
add a comment |
It returns Invalid syntax on df3.loc[(df3['Code_df1'] == df3['Code_df2']) & df3.loc[(df3['Owner_df1'] == df3['Owner_df2']),'select'] = 1
– gabriella
Nov 29 '18 at 8:50
Use this I missed a closing parenthesis while copying the code. df3.loc[(df3['Code_df1'] == df3['Code_df2']) & (df3.loc[(df3['Owner_df1'] == df3['Owner_df2']),'select'])] = 1
– Abhishek Sharma
Nov 30 '18 at 9:10
It returns Invalid syntax on df3.loc[(df3['Code_df1'] == df3['Code_df2']) & df3.loc[(df3['Owner_df1'] == df3['Owner_df2']),'select'] = 1
– gabriella
Nov 29 '18 at 8:50
It returns Invalid syntax on df3.loc[(df3['Code_df1'] == df3['Code_df2']) & df3.loc[(df3['Owner_df1'] == df3['Owner_df2']),'select'] = 1
– gabriella
Nov 29 '18 at 8:50
Use this I missed a closing parenthesis while copying the code. df3.loc[(df3['Code_df1'] == df3['Code_df2']) & (df3.loc[(df3['Owner_df1'] == df3['Owner_df2']),'select'])] = 1
– Abhishek Sharma
Nov 30 '18 at 9:10
Use this I missed a closing parenthesis while copying the code. df3.loc[(df3['Code_df1'] == df3['Code_df2']) & (df3.loc[(df3['Owner_df1'] == df3['Owner_df2']),'select'])] = 1
– Abhishek Sharma
Nov 30 '18 at 9:10
add a comment |
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I would think better is to have equally shaped series and try to use something like not s1.intersection(s2) or difference stackoverflow.com/questions/18079563/… pandas.pydata.org/pandas-docs/stable/generated/…
– Sergii
Nov 27 '18 at 12:18