Validate Pandas dataframe column based on hierarchy
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I have a dataframe like this
df1 = pd.DataFrame({'Site': ["S1", "S2", "S3", "S4", "S5", "S6","S7","S8","S9"],
'Sitelink': [" ","S1","S2","S6","S4"," ","S8"," ","S7"],
'level': ["R", "T", "P", "T", "P", "R","T","R","P"],
'Weight':["55","55","55","85","85","80","150","190","200"]})
column 'Site' will be always unique
column 'Sitelink' captures the next lower level site to each Site
column 'level' has 3 values- R, T, P where the hierarchy is R < T < P.
column 'Weight' can be any value.
The output should satisfy the condition that weight of a higher level site should be always lesser than or equal to lower level site. Expected result dataframe should be like
I'm trying to loop the dataframe and compare each site with next level. Is there a better approach to do this?
pandas dataframe
asked Nov 22 at 17:24
Osceria
479
add a comment |
up vote
0
down vote
favorite
I have a dataframe like this
df1 = pd.DataFrame({'Site': ["S1", "S2", "S3", "S4", "S5", "S6","S7","S8","S9"],
'Sitelink': [" ","S1","S2","S6","S4"," ","S8"," ","S7"],
'level': ["R", "T", "P", "T", "P", "R","T","R","P"],
'Weight':["55","55","55","85","85","80","150","190","200"]})
column 'Site' will be always unique
column 'Sitelink' captures the next lower level site to each Site
column 'level' has 3 values- R, T, P where the hierarchy is R < T < P.
column 'Weight' can be any value.
The output should satisfy the condition that weight of a higher level site should be always lesser than or equal to lower level site. Expected result dataframe should be like
I'm trying to loop the dataframe and compare each site with next level. Is there a better approach to do this?
pandas dataframe
asked Nov 22 at 17:24
Osceria
479
2
why row number 5 S5 S4 , value is equal , but return the error
– W-B
Nov 22 at 17:55
@W-B Yes, you're right. Corrected the question
– Osceria
Nov 22 at 22:42
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a dataframe like this
df1 = pd.DataFrame({'Site': ["S1", "S2", "S3", "S4", "S5", "S6","S7","S8","S9"],
'Sitelink': [" ","S1","S2","S6","S4"," ","S8"," ","S7"],
'level': ["R", "T", "P", "T", "P", "R","T","R","P"],
'Weight':["55","55","55","85","85","80","150","190","200"]})
column 'Site' will be always unique
column 'Sitelink' captures the next lower level site to each Site
column 'level' has 3 values- R, T, P where the hierarchy is R < T < P.
column 'Weight' can be any value.
The output should satisfy the condition that weight of a higher level site should be always lesser than or equal to lower level site. Expected result dataframe should be like
I'm trying to loop the dataframe and compare each site with next level. Is there a better approach to do this?
pandas dataframe
asked Nov 22 at 17:24
Osceria
479
I have a dataframe like this
df1 = pd.DataFrame({'Site': ["S1", "S2", "S3", "S4", "S5", "S6","S7","S8","S9"],
'Sitelink': [" ","S1","S2","S6","S4"," ","S8"," ","S7"],
'level': ["R", "T", "P", "T", "P", "R","T","R","P"],
'Weight':["55","55","55","85","85","80","150","190","200"]})
column 'Site' will be always unique
column 'Sitelink' captures the next lower level site to each Site
column 'level' has 3 values- R, T, P where the hierarchy is R < T < P.
column 'Weight' can be any value.
The output should satisfy the condition that weight of a higher level site should be always lesser than or equal to lower level site. Expected result dataframe should be like
I'm trying to loop the dataframe and compare each site with next level. Is there a better approach to do this?
pandas dataframe
pandas dataframe
asked Nov 22 at 17:24
Osceria
479
asked Nov 22 at 17:24
Osceria
479
edited Nov 22 at 22:41
asked Nov 22 at 17:24
Osceria
479
asked Nov 22 at 17:24
Osceria
479
asked Nov 22 at 17:24
Osceria
479
479
2
why row number 5 S5 S4 , value is equal , but return the error
– W-B
Nov 22 at 17:55
@W-B Yes, you're right. Corrected the question
– Osceria
Nov 22 at 22:42
add a comment |
2
why row number 5 S5 S4 , value is equal , but return the error
– W-B
Nov 22 at 17:55
@W-B Yes, you're right. Corrected the question
– Osceria
Nov 22 at 22:42
2
2
why row number 5 S5 S4 , value is equal , but return the error
– W-B
Nov 22 at 17:55
why row number 5 S5 S4 , value is equal , but return the error
– W-B
Nov 22 at 17:55
@W-B Yes, you're right. Corrected the question
– Osceria
Nov 22 at 22:42
@W-B Yes, you're right. Corrected the question
– Osceria
Nov 22 at 22:42
add a comment |
1 Answer
1
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votes
up vote
0
down vote
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If i understand correctly, you would like to to check if for each row the weight of the site is lower than or equal to the weight of the site marked as the Sitelink.
The code for a single row would than be:
def is_error(row):
if row['Sitelink'] == " ":
return 'No Error'
site_link = df.loc[df['Site'] == row['Sitelink']]
if int(row['Weight']) <= int(site_link['Weight']):
return 'No Error'
else:
return 'Higher than lower'
Therefore we could apply this line for each row using the apply function:
df['Error'] = df.apply(is_error, axis=1)
answered Nov 22 at 22:16
Gal Avineri
15210
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
If i understand correctly, you would like to to check if for each row the weight of the site is lower than or equal to the weight of the site marked as the Sitelink.
The code for a single row would than be:
def is_error(row):
if row['Sitelink'] == " ":
return 'No Error'
site_link = df.loc[df['Site'] == row['Sitelink']]
if int(row['Weight']) <= int(site_link['Weight']):
return 'No Error'
else:
return 'Higher than lower'
Therefore we could apply this line for each row using the apply function:
df['Error'] = df.apply(is_error, axis=1)
answered Nov 22 at 22:16
Gal Avineri
15210
add a comment |
up vote
0
down vote
accepted
If i understand correctly, you would like to to check if for each row the weight of the site is lower than or equal to the weight of the site marked as the Sitelink.
The code for a single row would than be:
def is_error(row):
if row['Sitelink'] == " ":
return 'No Error'
site_link = df.loc[df['Site'] == row['Sitelink']]
if int(row['Weight']) <= int(site_link['Weight']):
return 'No Error'
else:
return 'Higher than lower'
Therefore we could apply this line for each row using the apply function:
df['Error'] = df.apply(is_error, axis=1)
answered Nov 22 at 22:16
Gal Avineri
15210
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
If i understand correctly, you would like to to check if for each row the weight of the site is lower than or equal to the weight of the site marked as the Sitelink.
The code for a single row would than be:
def is_error(row):
if row['Sitelink'] == " ":
return 'No Error'
site_link = df.loc[df['Site'] == row['Sitelink']]
if int(row['Weight']) <= int(site_link['Weight']):
return 'No Error'
else:
return 'Higher than lower'
Therefore we could apply this line for each row using the apply function:
df['Error'] = df.apply(is_error, axis=1)
answered Nov 22 at 22:16
Gal Avineri
15210
If i understand correctly, you would like to to check if for each row the weight of the site is lower than or equal to the weight of the site marked as the Sitelink.
The code for a single row would than be:
def is_error(row):
if row['Sitelink'] == " ":
return 'No Error'
site_link = df.loc[df['Site'] == row['Sitelink']]
if int(row['Weight']) <= int(site_link['Weight']):
return 'No Error'
else:
return 'Higher than lower'
Therefore we could apply this line for each row using the apply function:
df['Error'] = df.apply(is_error, axis=1)
answered Nov 22 at 22:16
Gal Avineri
15210
edited Nov 22 at 23:07
answered Nov 22 at 22:16
Gal Avineri
15210
answered Nov 22 at 22:16
Gal Avineri
15210
answered Nov 22 at 22:16
Gal Avineri
15210
15210
add a comment |
add a comment |
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2
why row number 5 S5 S4 , value is equal , but return the error
– W-B
Nov 22 at 17:55
@W-B Yes, you're right. Corrected the question
– Osceria
Nov 22 at 22:42