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Get a count of combinations and its reverse from two columns

How to get count of two-way combinations from two columns?How to return multiple values from a function?Speed comparison with Project Euler: C vs Python vs Erlang vs HaskellPandas: create two new columns in a dataframe with values calculated from a pre-existing columnDelete column from pandas DataFrame by column name“Large data” work flows using pandasSelect rows from a DataFrame based on values in a column in pandasPandas DataFrame Groupby two columns and get countsGet list from pandas DataFrame column headerspandas create new column based on values from other columnsHow to count unique records by two columns in pandas?

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0

I'm trying to get a count of combinations from a pandas dataframe where it views reversed form of the combinations as the same. ie (A/B will be the same as B/A)

Similar to what this user is trying to do, but on python/pandas

How to get count of two-way combinations from two columns?

Thank you for helping!

I've explored crosstabs and grouping the data and it produces a count of the combinations, but it sees the reverse order as a unique combination.

Origin Destination
City 1 City 2
City 2 City 1
City 3 City 4
City 2 City 1

End result will look like

Route Count
City 1 - City 2 3
City 3 - City 4 1

note: order of the route does not matter. It could be City 2 - City 1, as long as it counts it as the same.

python pandas

share|improve this question

asked Mar 23 at 21:10

Tony NguyenTony Nguyen

1

add a comment | 

0

I'm trying to get a count of combinations from a pandas dataframe where it views reversed form of the combinations as the same. ie (A/B will be the same as B/A)

Similar to what this user is trying to do, but on python/pandas

How to get count of two-way combinations from two columns?

Thank you for helping!

I've explored crosstabs and grouping the data and it produces a count of the combinations, but it sees the reverse order as a unique combination.

Origin Destination
City 1 City 2
City 2 City 1
City 3 City 4
City 2 City 1

End result will look like

Route Count
City 1 - City 2 3
City 3 - City 4 1

note: order of the route does not matter. It could be City 2 - City 1, as long as it counts it as the same.

python pandas

share|improve this question

asked Mar 23 at 21:10

Tony NguyenTony Nguyen

1

add a comment | 

I'm trying to get a count of combinations from a pandas dataframe where it views reversed form of the combinations as the same. ie (A/B will be the same as B/A)

Similar to what this user is trying to do, but on python/pandas

How to get count of two-way combinations from two columns?

Thank you for helping!

I've explored crosstabs and grouping the data and it produces a count of the combinations, but it sees the reverse order as a unique combination.

Origin Destination
City 1 City 2
City 2 City 1
City 3 City 4
City 2 City 1

End result will look like

Route Count
City 1 - City 2 3
City 3 - City 4 1

note: order of the route does not matter. It could be City 2 - City 1, as long as it counts it as the same.

python pandas

share|improve this question

asked Mar 23 at 21:10

Tony NguyenTony Nguyen

1

Origin Destination
City 1 City 2
City 2 City 1
City 3 City 4
City 2 City 1

Route Count
City 1 - City 2 3
City 3 - City 4 1

share|improve this question

asked Mar 23 at 21:10

Tony NguyenTony Nguyen

1

share|improve this question

asked Mar 23 at 21:10

Tony NguyenTony Nguyen

1

asked Mar 23 at 21:10

Tony NguyenTony Nguyen

1

asked Mar 23 at 21:10

Tony NguyenTony Nguyen

1

2 Answers
2

active

oldest

votes

1

You can define a route using np.sort

import numpy as np
import pandas as pd

df['Route'] = [' - '.join(x) for x in np.sort(df.to_numpy(), axis=1)]
df.groupby('Route').size()

#Route
#City 1 - City 2 3
#City 3 - City 4 1
#dtype: int64

---

You can also construct a new sorted DataFrame, which could be useful:

df = pd.DataFrame(np.sort(df.to_numpy(), axis=1), index=df.index, columns=df.columns)

# Origin Destination
#0 City 1 City 2
#1 City 1 City 2
#2 City 3 City 4
#3 City 1 City 2

Now you can groupby ['Origin', 'Destintion']

share|improve this answer

answered Mar 23 at 21:23

ALollzALollz

18.5k51840

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Worked like a charm. Thank you Alollz. You are a GENIUS!!!!
  
  – Tony Nguyen
  Mar 23 at 22:39
  

  
  
  
  

add a comment | 

1

Check with sort

df.values.sort()
df.groupby(list(df)).size()
Origin Destination
City1 City2 3
City3 City4 1
dtype: int64

share|improve this answer

answered Mar 24 at 0:10

WeNYoBenWeNYoBen

135k84474

add a comment | 

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1

You can define a route using np.sort

import numpy as np
import pandas as pd

df['Route'] = [' - '.join(x) for x in np.sort(df.to_numpy(), axis=1)]
df.groupby('Route').size()

#Route
#City 1 - City 2 3
#City 3 - City 4 1
#dtype: int64

---

You can also construct a new sorted DataFrame, which could be useful:

df = pd.DataFrame(np.sort(df.to_numpy(), axis=1), index=df.index, columns=df.columns)

# Origin Destination
#0 City 1 City 2
#1 City 1 City 2
#2 City 3 City 4
#3 City 1 City 2

Now you can groupby ['Origin', 'Destintion']

share|improve this answer

answered Mar 23 at 21:23

ALollzALollz

18.5k51840

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Worked like a charm. Thank you Alollz. You are a GENIUS!!!!
  
  – Tony Nguyen
  Mar 23 at 22:39
  

  
  
  
  

add a comment | 

1

You can define a route using np.sort

import numpy as np
import pandas as pd

df['Route'] = [' - '.join(x) for x in np.sort(df.to_numpy(), axis=1)]
df.groupby('Route').size()

#Route
#City 1 - City 2 3
#City 3 - City 4 1
#dtype: int64

---

You can also construct a new sorted DataFrame, which could be useful:

df = pd.DataFrame(np.sort(df.to_numpy(), axis=1), index=df.index, columns=df.columns)

# Origin Destination
#0 City 1 City 2
#1 City 1 City 2
#2 City 3 City 4
#3 City 1 City 2

Now you can groupby ['Origin', 'Destintion']

share|improve this answer

answered Mar 23 at 21:23

ALollzALollz

18.5k51840

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Worked like a charm. Thank you Alollz. You are a GENIUS!!!!
  
  – Tony Nguyen
  Mar 23 at 22:39
  

  
  
  
  

add a comment | 

You can define a route using np.sort

import numpy as np
import pandas as pd

df['Route'] = [' - '.join(x) for x in np.sort(df.to_numpy(), axis=1)]
df.groupby('Route').size()

#Route
#City 1 - City 2 3
#City 3 - City 4 1
#dtype: int64

---

You can also construct a new sorted DataFrame, which could be useful:

df = pd.DataFrame(np.sort(df.to_numpy(), axis=1), index=df.index, columns=df.columns)

# Origin Destination
#0 City 1 City 2
#1 City 1 City 2
#2 City 3 City 4
#3 City 1 City 2

Now you can groupby ['Origin', 'Destintion']

share|improve this answer

answered Mar 23 at 21:23

ALollzALollz

18.5k51840

import numpy as np
import pandas as pd

df['Route'] = [' - '.join(x) for x in np.sort(df.to_numpy(), axis=1)]
df.groupby('Route').size()

#Route
#City 1 - City 2 3
#City 3 - City 4 1
#dtype: int64

df = pd.DataFrame(np.sort(df.to_numpy(), axis=1), index=df.index, columns=df.columns)

# Origin Destination
#0 City 1 City 2
#1 City 1 City 2
#2 City 3 City 4
#3 City 1 City 2

share|improve this answer

answered Mar 23 at 21:23

ALollzALollz

18.5k51840

answered Mar 23 at 21:23

ALollzALollz

18.5k51840

answered Mar 23 at 21:23

ALollzALollz

18.5k51840

1

Check with sort

df.values.sort()
df.groupby(list(df)).size()
Origin Destination
City1 City2 3
City3 City4 1
dtype: int64

share|improve this answer

answered Mar 24 at 0:10

WeNYoBenWeNYoBen

135k84474

add a comment | 

1

Check with sort

df.values.sort()
df.groupby(list(df)).size()
Origin Destination
City1 City2 3
City3 City4 1
dtype: int64

share|improve this answer

answered Mar 24 at 0:10

WeNYoBenWeNYoBen

135k84474

add a comment | 

Check with sort

df.values.sort()
df.groupby(list(df)).size()
Origin Destination
City1 City2 3
City3 City4 1
dtype: int64

share|improve this answer

answered Mar 24 at 0:10

WeNYoBenWeNYoBen

135k84474

df.values.sort()
df.groupby(list(df)).size()
Origin Destination
City1 City2 3
City3 City4 1
dtype: int64

share|improve this answer

answered Mar 24 at 0:10

WeNYoBenWeNYoBen

135k84474

answered Mar 24 at 0:10

WeNYoBenWeNYoBen

135k84474

answered Mar 24 at 0:10

WeNYoBenWeNYoBen

135k84474