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2

I am attempting to label accounts as new, current, lost, or returning but am having trouble with the logic. The row index is the account and the columns are the years and the values are 1's and 0's representing if the account is active or not. This is what i came up with so far. I'm not sure if this will ever work or if i'm close and I'm not sure how the logic would look for returning customers.
df2 is the original data frame and df3 = df2.shift(periods=1,axis=1)

def differences():
 if df2 != df3 & df2 == 1:
 return "New"
 elif df2 != df3 & df2 ==0:
 return "Lost"
 elif df2 == df3 & df2 ==0:
 return ""
 else:
 return "Continuing"
differences() 

`

And when I run this code i get the following error:

couldn't find matching opcode for 'and_bdl'

python-3.x pandas jupyter-notebook

share|improve this question

asked Mar 23 at 1:42

jwrigjwrig

133

add a comment | 

2

I am attempting to label accounts as new, current, lost, or returning but am having trouble with the logic. The row index is the account and the columns are the years and the values are 1's and 0's representing if the account is active or not. This is what i came up with so far. I'm not sure if this will ever work or if i'm close and I'm not sure how the logic would look for returning customers.
df2 is the original data frame and df3 = df2.shift(periods=1,axis=1)

def differences():
 if df2 != df3 & df2 == 1:
 return "New"
 elif df2 != df3 & df2 ==0:
 return "Lost"
 elif df2 == df3 & df2 ==0:
 return ""
 else:
 return "Continuing"
differences() 

`

And when I run this code i get the following error:

couldn't find matching opcode for 'and_bdl'

python-3.x pandas jupyter-notebook

share|improve this question

asked Mar 23 at 1:42

jwrigjwrig

133

add a comment | 

I am attempting to label accounts as new, current, lost, or returning but am having trouble with the logic. The row index is the account and the columns are the years and the values are 1's and 0's representing if the account is active or not. This is what i came up with so far. I'm not sure if this will ever work or if i'm close and I'm not sure how the logic would look for returning customers.
df2 is the original data frame and df3 = df2.shift(periods=1,axis=1)

def differences():
 if df2 != df3 & df2 == 1:
 return "New"
 elif df2 != df3 & df2 ==0:
 return "Lost"
 elif df2 == df3 & df2 ==0:
 return ""
 else:
 return "Continuing"
differences() 

`

And when I run this code i get the following error:

couldn't find matching opcode for 'and_bdl'

python-3.x pandas jupyter-notebook

share|improve this question

asked Mar 23 at 1:42

jwrigjwrig

133

def differences():
 if df2 != df3 & df2 == 1:
 return "New"
 elif df2 != df3 & df2 ==0:
 return "Lost"
 elif df2 == df3 & df2 ==0:
 return ""
 else:
 return "Continuing"
differences() 

couldn't find matching opcode for 'and_bdl'

share|improve this question

asked Mar 23 at 1:42

jwrigjwrig

133

share|improve this question

asked Mar 23 at 1:42

jwrigjwrig

133

asked Mar 23 at 1:42

jwrigjwrig

133

asked Mar 23 at 1:42

jwrigjwrig

133

1 Answer
1

active

oldest

votes

1

The following code logic might work for you case.

EDIT: Based on your comment, I modified the code so that all columns except the last one are checked.

import pandas as pd

str="""account 2019 2018 2017 2016 2015
alex 1 0 0 0 0
joe 0 0 1 0 0
boss 1 1 1 1 1
smith 1 1 0 1 0"""
df = pd.read_csv(pd.io.common.StringIO(str), sep='s+', index_col='account')
df
#Out[46]: 
# 2019 2018 2017 2016 2015
#account 
#alex 1 0 0 0 0
#joe 0 0 1 0 0
#boss 1 1 1 1 1
#smith 1 1 0 1 0

# find account status per-year
def account_status(x):
 status = []
 n = x.size
 for i in range(n-1):
 if x.iloc[i] == 1: 
 # if all rest are '0'
 if x.iloc[i+1:].eq(0).all():
 status.extend(['new'] + [None]*(n-i-2))
 break 
 # if the previous year is '0'
 elif x.iloc[i+1] == 0:
 status.append('returning')
 else:
 status.append('continuing')
 else:
 # at least one '1' in previous years
 if x.iloc[i+1:].eq(1).any():
 status.append('lost')
 else:
 status.extend([None] * (n-i-1))
 break
 return status 

s = df.apply(account_status, axis=1).apply(pd.Series)
s.columns = df.columns[:-1]
s 
#Out[57]: 
# 2019 2018 2017 2016
#account 
#alex new None None None
#joe lost lost new None
#boss continuing continuing continuing continuing
#smith continuing returning lost new

share|improve this answer

edited Mar 23 at 15:28

answered Mar 23 at 3:56

jxcjxc

1,5232310

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you very much for your response! That code makes a lot of sense. would there be a way to tweak it to make it so that it returns a value for every column? For instance return #smith continuing returning lost new " "
  
  – jwrig
  Mar 23 at 5:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  updated per your comment.
  
  – jxc
  Mar 23 at 15:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That is exactly the result i was looking for. Thank you very much for your effort, patience and insight. I'm very new to python so you're showing me a lot of things that I will be looking deeper into as i expect to be using pandas regularly.
  
  – jwrig
  Mar 24 at 2:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  glad it helped and good luck to you with the new journey.
  
  – jxc
  Mar 24 at 2:59
  

  
  
  
  

add a comment | 

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1

The following code logic might work for you case.

EDIT: Based on your comment, I modified the code so that all columns except the last one are checked.

import pandas as pd

str="""account 2019 2018 2017 2016 2015
alex 1 0 0 0 0
joe 0 0 1 0 0
boss 1 1 1 1 1
smith 1 1 0 1 0"""
df = pd.read_csv(pd.io.common.StringIO(str), sep='s+', index_col='account')
df
#Out[46]: 
# 2019 2018 2017 2016 2015
#account 
#alex 1 0 0 0 0
#joe 0 0 1 0 0
#boss 1 1 1 1 1
#smith 1 1 0 1 0

# find account status per-year
def account_status(x):
 status = []
 n = x.size
 for i in range(n-1):
 if x.iloc[i] == 1: 
 # if all rest are '0'
 if x.iloc[i+1:].eq(0).all():
 status.extend(['new'] + [None]*(n-i-2))
 break 
 # if the previous year is '0'
 elif x.iloc[i+1] == 0:
 status.append('returning')
 else:
 status.append('continuing')
 else:
 # at least one '1' in previous years
 if x.iloc[i+1:].eq(1).any():
 status.append('lost')
 else:
 status.extend([None] * (n-i-1))
 break
 return status 

s = df.apply(account_status, axis=1).apply(pd.Series)
s.columns = df.columns[:-1]
s 
#Out[57]: 
# 2019 2018 2017 2016
#account 
#alex new None None None
#joe lost lost new None
#boss continuing continuing continuing continuing
#smith continuing returning lost new

share|improve this answer

edited Mar 23 at 15:28

answered Mar 23 at 3:56

jxcjxc

1,5232310

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you very much for your response! That code makes a lot of sense. would there be a way to tweak it to make it so that it returns a value for every column? For instance return #smith continuing returning lost new " "
  
  – jwrig
  Mar 23 at 5:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  updated per your comment.
  
  – jxc
  Mar 23 at 15:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That is exactly the result i was looking for. Thank you very much for your effort, patience and insight. I'm very new to python so you're showing me a lot of things that I will be looking deeper into as i expect to be using pandas regularly.
  
  – jwrig
  Mar 24 at 2:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  glad it helped and good luck to you with the new journey.
  
  – jxc
  Mar 24 at 2:59
  

  
  
  
  

add a comment | 

1

The following code logic might work for you case.

EDIT: Based on your comment, I modified the code so that all columns except the last one are checked.

import pandas as pd

str="""account 2019 2018 2017 2016 2015
alex 1 0 0 0 0
joe 0 0 1 0 0
boss 1 1 1 1 1
smith 1 1 0 1 0"""
df = pd.read_csv(pd.io.common.StringIO(str), sep='s+', index_col='account')
df
#Out[46]: 
# 2019 2018 2017 2016 2015
#account 
#alex 1 0 0 0 0
#joe 0 0 1 0 0
#boss 1 1 1 1 1
#smith 1 1 0 1 0

# find account status per-year
def account_status(x):
 status = []
 n = x.size
 for i in range(n-1):
 if x.iloc[i] == 1: 
 # if all rest are '0'
 if x.iloc[i+1:].eq(0).all():
 status.extend(['new'] + [None]*(n-i-2))
 break 
 # if the previous year is '0'
 elif x.iloc[i+1] == 0:
 status.append('returning')
 else:
 status.append('continuing')
 else:
 # at least one '1' in previous years
 if x.iloc[i+1:].eq(1).any():
 status.append('lost')
 else:
 status.extend([None] * (n-i-1))
 break
 return status 

s = df.apply(account_status, axis=1).apply(pd.Series)
s.columns = df.columns[:-1]
s 
#Out[57]: 
# 2019 2018 2017 2016
#account 
#alex new None None None
#joe lost lost new None
#boss continuing continuing continuing continuing
#smith continuing returning lost new

share|improve this answer

edited Mar 23 at 15:28

answered Mar 23 at 3:56

jxcjxc

1,5232310

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you very much for your response! That code makes a lot of sense. would there be a way to tweak it to make it so that it returns a value for every column? For instance return #smith continuing returning lost new " "
  
  – jwrig
  Mar 23 at 5:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  updated per your comment.
  
  – jxc
  Mar 23 at 15:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That is exactly the result i was looking for. Thank you very much for your effort, patience and insight. I'm very new to python so you're showing me a lot of things that I will be looking deeper into as i expect to be using pandas regularly.
  
  – jwrig
  Mar 24 at 2:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  glad it helped and good luck to you with the new journey.
  
  – jxc
  Mar 24 at 2:59
  

  
  
  
  

add a comment | 

The following code logic might work for you case.

EDIT: Based on your comment, I modified the code so that all columns except the last one are checked.

import pandas as pd

str="""account 2019 2018 2017 2016 2015
alex 1 0 0 0 0
joe 0 0 1 0 0
boss 1 1 1 1 1
smith 1 1 0 1 0"""
df = pd.read_csv(pd.io.common.StringIO(str), sep='s+', index_col='account')
df
#Out[46]: 
# 2019 2018 2017 2016 2015
#account 
#alex 1 0 0 0 0
#joe 0 0 1 0 0
#boss 1 1 1 1 1
#smith 1 1 0 1 0

# find account status per-year
def account_status(x):
 status = []
 n = x.size
 for i in range(n-1):
 if x.iloc[i] == 1: 
 # if all rest are '0'
 if x.iloc[i+1:].eq(0).all():
 status.extend(['new'] + [None]*(n-i-2))
 break 
 # if the previous year is '0'
 elif x.iloc[i+1] == 0:
 status.append('returning')
 else:
 status.append('continuing')
 else:
 # at least one '1' in previous years
 if x.iloc[i+1:].eq(1).any():
 status.append('lost')
 else:
 status.extend([None] * (n-i-1))
 break
 return status 

s = df.apply(account_status, axis=1).apply(pd.Series)
s.columns = df.columns[:-1]
s 
#Out[57]: 
# 2019 2018 2017 2016
#account 
#alex new None None None
#joe lost lost new None
#boss continuing continuing continuing continuing
#smith continuing returning lost new

share|improve this answer

edited Mar 23 at 15:28

answered Mar 23 at 3:56

jxcjxc

1,5232310

import pandas as pd

str="""account 2019 2018 2017 2016 2015
alex 1 0 0 0 0
joe 0 0 1 0 0
boss 1 1 1 1 1
smith 1 1 0 1 0"""
df = pd.read_csv(pd.io.common.StringIO(str), sep='s+', index_col='account')
df
#Out[46]: 
# 2019 2018 2017 2016 2015
#account 
#alex 1 0 0 0 0
#joe 0 0 1 0 0
#boss 1 1 1 1 1
#smith 1 1 0 1 0

# find account status per-year
def account_status(x):
 status = []
 n = x.size
 for i in range(n-1):
 if x.iloc[i] == 1: 
 # if all rest are '0'
 if x.iloc[i+1:].eq(0).all():
 status.extend(['new'] + [None]*(n-i-2))
 break 
 # if the previous year is '0'
 elif x.iloc[i+1] == 0:
 status.append('returning')
 else:
 status.append('continuing')
 else:
 # at least one '1' in previous years
 if x.iloc[i+1:].eq(1).any():
 status.append('lost')
 else:
 status.extend([None] * (n-i-1))
 break
 return status 

s = df.apply(account_status, axis=1).apply(pd.Series)
s.columns = df.columns[:-1]
s 
#Out[57]: 
# 2019 2018 2017 2016
#account 
#alex new None None None
#joe lost lost new None
#boss continuing continuing continuing continuing
#smith continuing returning lost new

share|improve this answer

edited Mar 23 at 15:28

answered Mar 23 at 3:56

jxcjxc

1,5232310

answered Mar 23 at 3:56

jxcjxc

1,5232310

answered Mar 23 at 3:56

jxcjxc

1,5232310