Python pandas - add multiple columns to dataframe where rows and columns are co-dependent










1















I have managed to get this working with for-loops but these are very slow on the large datasets I'm working with, so am trying to find a way to do this using pandas, groupby, apply and lamda functions instead.



import pandas as pd
example_df = pd.DataFrame("scen": [1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2],
"cusip": ['031162CF5', '031162CF5', '031162CF5', '031162CF5', '38141GWM2', '38141GWM2', '38141GWM2', '38141GWM2', '031162CF5', '031162CF5', '031162CF5', '031162CF5', '38141GWM2', '38141GWM2', '38141GWM2', '38141GWM2'],
"wal": [50, 55, 60, 65, 40, 50, 60, 70, 40, 45, 50, 55, 30, 40, 50, 60],
"par_val": [900000, 800000, 700000, 600000, 900000, 800000, 700000, 600000, 900000, 800000, 700000, 600000, 900000, 800000, 700000, 600000],
"prin_cf": [0, 100000, 100000, 100000, 0, 100000, 100000, 100000, 0, 100000, 100000, 100000, 0, 100000, 100000, 100000],
"amortization": [166.67, 0, 0, 0, 208.33, 0, 0, 0, 208.33, 0, 0, 0, 277.78, 0, 0, 0],
"book_val": [1000000, 0, 0, 0, 1000000, 0, 0, 0, 1000000, 0, 0, 0, 1000000, 0, 0, 0])

for x in range(1, len(example_df['scen'])):

if (example_df['cusip'][x] == example_df['cusip'][x-1]):

# If bond matures, don't report book value
if(example_df['par_val'][x] == 0):
example_df['book_val'][x] = 0
else:
example_df['book_val'][x] = example_df['book_val'][x-1] - example_df['amortization'][x-1] - example_df['prin_cf'][x-1]


example_df['amortization'][x] = (example_df['book_val'][x] - example_df['par_val'][x]) / example_df['wal'][x] / 12

example_df


The tricky part is that each row's book value depends on the previous row's amortization value, while each amortization value depends on the book value in the same row. Looking at the responses to a similar question here, I think there may be a way to do this using global variables that keep track of the previous values.



Is there a way in Pandas to use previous row value in dataframe.apply when previous value is also calculated in the apply?



Something like:



def calc_bv(prin_cf, par_val, wal):
global bvalue, amort
bvalue = bvalue - amort - prin_cf
amort = (bvalue - par_val)/wal/12
return bvalue, amort

bvalue = example_df.loc[0, 'book_val']
amort = example_df.loc[0, 'amortization']
example_df[1:][['book_val','amortization']] = example_df2[1:].apply(lambda row: calc_bv(row['prev_prin_cf'],row['par_val'],row['wal']), axis=1, result_type="expand")
example_df









share|improve this question




























    1















    I have managed to get this working with for-loops but these are very slow on the large datasets I'm working with, so am trying to find a way to do this using pandas, groupby, apply and lamda functions instead.



    import pandas as pd
    example_df = pd.DataFrame("scen": [1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2],
    "cusip": ['031162CF5', '031162CF5', '031162CF5', '031162CF5', '38141GWM2', '38141GWM2', '38141GWM2', '38141GWM2', '031162CF5', '031162CF5', '031162CF5', '031162CF5', '38141GWM2', '38141GWM2', '38141GWM2', '38141GWM2'],
    "wal": [50, 55, 60, 65, 40, 50, 60, 70, 40, 45, 50, 55, 30, 40, 50, 60],
    "par_val": [900000, 800000, 700000, 600000, 900000, 800000, 700000, 600000, 900000, 800000, 700000, 600000, 900000, 800000, 700000, 600000],
    "prin_cf": [0, 100000, 100000, 100000, 0, 100000, 100000, 100000, 0, 100000, 100000, 100000, 0, 100000, 100000, 100000],
    "amortization": [166.67, 0, 0, 0, 208.33, 0, 0, 0, 208.33, 0, 0, 0, 277.78, 0, 0, 0],
    "book_val": [1000000, 0, 0, 0, 1000000, 0, 0, 0, 1000000, 0, 0, 0, 1000000, 0, 0, 0])

    for x in range(1, len(example_df['scen'])):

    if (example_df['cusip'][x] == example_df['cusip'][x-1]):

    # If bond matures, don't report book value
    if(example_df['par_val'][x] == 0):
    example_df['book_val'][x] = 0
    else:
    example_df['book_val'][x] = example_df['book_val'][x-1] - example_df['amortization'][x-1] - example_df['prin_cf'][x-1]


    example_df['amortization'][x] = (example_df['book_val'][x] - example_df['par_val'][x]) / example_df['wal'][x] / 12

    example_df


    The tricky part is that each row's book value depends on the previous row's amortization value, while each amortization value depends on the book value in the same row. Looking at the responses to a similar question here, I think there may be a way to do this using global variables that keep track of the previous values.



    Is there a way in Pandas to use previous row value in dataframe.apply when previous value is also calculated in the apply?



    Something like:



    def calc_bv(prin_cf, par_val, wal):
    global bvalue, amort
    bvalue = bvalue - amort - prin_cf
    amort = (bvalue - par_val)/wal/12
    return bvalue, amort

    bvalue = example_df.loc[0, 'book_val']
    amort = example_df.loc[0, 'amortization']
    example_df[1:][['book_val','amortization']] = example_df2[1:].apply(lambda row: calc_bv(row['prev_prin_cf'],row['par_val'],row['wal']), axis=1, result_type="expand")
    example_df









    share|improve this question


























      1












      1








      1








      I have managed to get this working with for-loops but these are very slow on the large datasets I'm working with, so am trying to find a way to do this using pandas, groupby, apply and lamda functions instead.



      import pandas as pd
      example_df = pd.DataFrame("scen": [1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2],
      "cusip": ['031162CF5', '031162CF5', '031162CF5', '031162CF5', '38141GWM2', '38141GWM2', '38141GWM2', '38141GWM2', '031162CF5', '031162CF5', '031162CF5', '031162CF5', '38141GWM2', '38141GWM2', '38141GWM2', '38141GWM2'],
      "wal": [50, 55, 60, 65, 40, 50, 60, 70, 40, 45, 50, 55, 30, 40, 50, 60],
      "par_val": [900000, 800000, 700000, 600000, 900000, 800000, 700000, 600000, 900000, 800000, 700000, 600000, 900000, 800000, 700000, 600000],
      "prin_cf": [0, 100000, 100000, 100000, 0, 100000, 100000, 100000, 0, 100000, 100000, 100000, 0, 100000, 100000, 100000],
      "amortization": [166.67, 0, 0, 0, 208.33, 0, 0, 0, 208.33, 0, 0, 0, 277.78, 0, 0, 0],
      "book_val": [1000000, 0, 0, 0, 1000000, 0, 0, 0, 1000000, 0, 0, 0, 1000000, 0, 0, 0])

      for x in range(1, len(example_df['scen'])):

      if (example_df['cusip'][x] == example_df['cusip'][x-1]):

      # If bond matures, don't report book value
      if(example_df['par_val'][x] == 0):
      example_df['book_val'][x] = 0
      else:
      example_df['book_val'][x] = example_df['book_val'][x-1] - example_df['amortization'][x-1] - example_df['prin_cf'][x-1]


      example_df['amortization'][x] = (example_df['book_val'][x] - example_df['par_val'][x]) / example_df['wal'][x] / 12

      example_df


      The tricky part is that each row's book value depends on the previous row's amortization value, while each amortization value depends on the book value in the same row. Looking at the responses to a similar question here, I think there may be a way to do this using global variables that keep track of the previous values.



      Is there a way in Pandas to use previous row value in dataframe.apply when previous value is also calculated in the apply?



      Something like:



      def calc_bv(prin_cf, par_val, wal):
      global bvalue, amort
      bvalue = bvalue - amort - prin_cf
      amort = (bvalue - par_val)/wal/12
      return bvalue, amort

      bvalue = example_df.loc[0, 'book_val']
      amort = example_df.loc[0, 'amortization']
      example_df[1:][['book_val','amortization']] = example_df2[1:].apply(lambda row: calc_bv(row['prev_prin_cf'],row['par_val'],row['wal']), axis=1, result_type="expand")
      example_df









      share|improve this question
















      I have managed to get this working with for-loops but these are very slow on the large datasets I'm working with, so am trying to find a way to do this using pandas, groupby, apply and lamda functions instead.



      import pandas as pd
      example_df = pd.DataFrame("scen": [1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2],
      "cusip": ['031162CF5', '031162CF5', '031162CF5', '031162CF5', '38141GWM2', '38141GWM2', '38141GWM2', '38141GWM2', '031162CF5', '031162CF5', '031162CF5', '031162CF5', '38141GWM2', '38141GWM2', '38141GWM2', '38141GWM2'],
      "wal": [50, 55, 60, 65, 40, 50, 60, 70, 40, 45, 50, 55, 30, 40, 50, 60],
      "par_val": [900000, 800000, 700000, 600000, 900000, 800000, 700000, 600000, 900000, 800000, 700000, 600000, 900000, 800000, 700000, 600000],
      "prin_cf": [0, 100000, 100000, 100000, 0, 100000, 100000, 100000, 0, 100000, 100000, 100000, 0, 100000, 100000, 100000],
      "amortization": [166.67, 0, 0, 0, 208.33, 0, 0, 0, 208.33, 0, 0, 0, 277.78, 0, 0, 0],
      "book_val": [1000000, 0, 0, 0, 1000000, 0, 0, 0, 1000000, 0, 0, 0, 1000000, 0, 0, 0])

      for x in range(1, len(example_df['scen'])):

      if (example_df['cusip'][x] == example_df['cusip'][x-1]):

      # If bond matures, don't report book value
      if(example_df['par_val'][x] == 0):
      example_df['book_val'][x] = 0
      else:
      example_df['book_val'][x] = example_df['book_val'][x-1] - example_df['amortization'][x-1] - example_df['prin_cf'][x-1]


      example_df['amortization'][x] = (example_df['book_val'][x] - example_df['par_val'][x]) / example_df['wal'][x] / 12

      example_df


      The tricky part is that each row's book value depends on the previous row's amortization value, while each amortization value depends on the book value in the same row. Looking at the responses to a similar question here, I think there may be a way to do this using global variables that keep track of the previous values.



      Is there a way in Pandas to use previous row value in dataframe.apply when previous value is also calculated in the apply?



      Something like:



      def calc_bv(prin_cf, par_val, wal):
      global bvalue, amort
      bvalue = bvalue - amort - prin_cf
      amort = (bvalue - par_val)/wal/12
      return bvalue, amort

      bvalue = example_df.loc[0, 'book_val']
      amort = example_df.loc[0, 'amortization']
      example_df[1:][['book_val','amortization']] = example_df2[1:].apply(lambda row: calc_bv(row['prev_prin_cf'],row['par_val'],row['wal']), axis=1, result_type="expand")
      example_df






      python pandas performance dataframe






      share|improve this question















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      share|improve this question




      share|improve this question








      edited Nov 13 '18 at 13:20









      Malik Asad

      296111




      296111










      asked Nov 13 '18 at 11:48









      M.HopeM.Hope

      112




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          1 Answer
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          0














          There will, no doubt, be a smart Pandas solution based on groupby. But you can get a decent, around ~1000x, performance improvement by just rewriting your loop with numba.



          # Python 3.6.0, Pandas 0.19.2

          assert jpp(df).equals(original(df))

          %timeit jpp(df) # 929 µs per loop
          %timeit original(df) # 1.05 s per loop


          Benchmarking code



          Original:



          def original(example_df):
          for x in range(1, len(example_df['scen'])):

          if (example_df['cusip'][x] == example_df['cusip'][x-1]):

          # If bond matures, don't report book value
          if(example_df['par_val'][x] == 0):
          example_df['book_val'][x] = 0
          else:
          example_df['book_val'][x] = example_df['book_val'][x-1] - example_df['amortization'][x-1] - example_df['prin_cf'][x-1]


          example_df['amortization'][x] = (example_df['book_val'][x] - example_df['par_val'][x]) / example_df['wal'][x] / 12
          return example_df


          Numba:



          from numba import njit

          @njit
          def calculator(cusip, par, book, amort, prin_cf, wal):
          n = len(par)
          for i in range(1, n):
          if cusip[i] == cusip[i-1]:
          if par[i] == 0:
          book[i] == 0
          else:
          book[i] = book[i-1] - amort[i-1] - prin_cf[i-1]
          amort[i] = (book[i] - par[i]) / wal[i] / 12
          return book, amort


          def jpp(df):
          df['book_val'], df['amortization'] = calculator(pd.factorize(df['cusip'])[0], df['par_val'].values,
          df['book_val'].values, df['amortization'].values,
          df['prin_cf'].values, df['wal'].values)

          return df





          share|improve this answer























          • Thanks! Although unfortunately numba is not on the list of libraries I have access to at work... any ideas on how the pandas/groupby solution might look?

            – M.Hope
            Dec 10 '18 at 15:29











          • @M.Hope, Sorry, can't think of anything. Get your work to install numba :)

            – jpp
            Dec 10 '18 at 15:35











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          1 Answer
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          1 Answer
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          active

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          active

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          active

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          0














          There will, no doubt, be a smart Pandas solution based on groupby. But you can get a decent, around ~1000x, performance improvement by just rewriting your loop with numba.



          # Python 3.6.0, Pandas 0.19.2

          assert jpp(df).equals(original(df))

          %timeit jpp(df) # 929 µs per loop
          %timeit original(df) # 1.05 s per loop


          Benchmarking code



          Original:



          def original(example_df):
          for x in range(1, len(example_df['scen'])):

          if (example_df['cusip'][x] == example_df['cusip'][x-1]):

          # If bond matures, don't report book value
          if(example_df['par_val'][x] == 0):
          example_df['book_val'][x] = 0
          else:
          example_df['book_val'][x] = example_df['book_val'][x-1] - example_df['amortization'][x-1] - example_df['prin_cf'][x-1]


          example_df['amortization'][x] = (example_df['book_val'][x] - example_df['par_val'][x]) / example_df['wal'][x] / 12
          return example_df


          Numba:



          from numba import njit

          @njit
          def calculator(cusip, par, book, amort, prin_cf, wal):
          n = len(par)
          for i in range(1, n):
          if cusip[i] == cusip[i-1]:
          if par[i] == 0:
          book[i] == 0
          else:
          book[i] = book[i-1] - amort[i-1] - prin_cf[i-1]
          amort[i] = (book[i] - par[i]) / wal[i] / 12
          return book, amort


          def jpp(df):
          df['book_val'], df['amortization'] = calculator(pd.factorize(df['cusip'])[0], df['par_val'].values,
          df['book_val'].values, df['amortization'].values,
          df['prin_cf'].values, df['wal'].values)

          return df





          share|improve this answer























          • Thanks! Although unfortunately numba is not on the list of libraries I have access to at work... any ideas on how the pandas/groupby solution might look?

            – M.Hope
            Dec 10 '18 at 15:29











          • @M.Hope, Sorry, can't think of anything. Get your work to install numba :)

            – jpp
            Dec 10 '18 at 15:35
















          0














          There will, no doubt, be a smart Pandas solution based on groupby. But you can get a decent, around ~1000x, performance improvement by just rewriting your loop with numba.



          # Python 3.6.0, Pandas 0.19.2

          assert jpp(df).equals(original(df))

          %timeit jpp(df) # 929 µs per loop
          %timeit original(df) # 1.05 s per loop


          Benchmarking code



          Original:



          def original(example_df):
          for x in range(1, len(example_df['scen'])):

          if (example_df['cusip'][x] == example_df['cusip'][x-1]):

          # If bond matures, don't report book value
          if(example_df['par_val'][x] == 0):
          example_df['book_val'][x] = 0
          else:
          example_df['book_val'][x] = example_df['book_val'][x-1] - example_df['amortization'][x-1] - example_df['prin_cf'][x-1]


          example_df['amortization'][x] = (example_df['book_val'][x] - example_df['par_val'][x]) / example_df['wal'][x] / 12
          return example_df


          Numba:



          from numba import njit

          @njit
          def calculator(cusip, par, book, amort, prin_cf, wal):
          n = len(par)
          for i in range(1, n):
          if cusip[i] == cusip[i-1]:
          if par[i] == 0:
          book[i] == 0
          else:
          book[i] = book[i-1] - amort[i-1] - prin_cf[i-1]
          amort[i] = (book[i] - par[i]) / wal[i] / 12
          return book, amort


          def jpp(df):
          df['book_val'], df['amortization'] = calculator(pd.factorize(df['cusip'])[0], df['par_val'].values,
          df['book_val'].values, df['amortization'].values,
          df['prin_cf'].values, df['wal'].values)

          return df





          share|improve this answer























          • Thanks! Although unfortunately numba is not on the list of libraries I have access to at work... any ideas on how the pandas/groupby solution might look?

            – M.Hope
            Dec 10 '18 at 15:29











          • @M.Hope, Sorry, can't think of anything. Get your work to install numba :)

            – jpp
            Dec 10 '18 at 15:35














          0












          0








          0







          There will, no doubt, be a smart Pandas solution based on groupby. But you can get a decent, around ~1000x, performance improvement by just rewriting your loop with numba.



          # Python 3.6.0, Pandas 0.19.2

          assert jpp(df).equals(original(df))

          %timeit jpp(df) # 929 µs per loop
          %timeit original(df) # 1.05 s per loop


          Benchmarking code



          Original:



          def original(example_df):
          for x in range(1, len(example_df['scen'])):

          if (example_df['cusip'][x] == example_df['cusip'][x-1]):

          # If bond matures, don't report book value
          if(example_df['par_val'][x] == 0):
          example_df['book_val'][x] = 0
          else:
          example_df['book_val'][x] = example_df['book_val'][x-1] - example_df['amortization'][x-1] - example_df['prin_cf'][x-1]


          example_df['amortization'][x] = (example_df['book_val'][x] - example_df['par_val'][x]) / example_df['wal'][x] / 12
          return example_df


          Numba:



          from numba import njit

          @njit
          def calculator(cusip, par, book, amort, prin_cf, wal):
          n = len(par)
          for i in range(1, n):
          if cusip[i] == cusip[i-1]:
          if par[i] == 0:
          book[i] == 0
          else:
          book[i] = book[i-1] - amort[i-1] - prin_cf[i-1]
          amort[i] = (book[i] - par[i]) / wal[i] / 12
          return book, amort


          def jpp(df):
          df['book_val'], df['amortization'] = calculator(pd.factorize(df['cusip'])[0], df['par_val'].values,
          df['book_val'].values, df['amortization'].values,
          df['prin_cf'].values, df['wal'].values)

          return df





          share|improve this answer













          There will, no doubt, be a smart Pandas solution based on groupby. But you can get a decent, around ~1000x, performance improvement by just rewriting your loop with numba.



          # Python 3.6.0, Pandas 0.19.2

          assert jpp(df).equals(original(df))

          %timeit jpp(df) # 929 µs per loop
          %timeit original(df) # 1.05 s per loop


          Benchmarking code



          Original:



          def original(example_df):
          for x in range(1, len(example_df['scen'])):

          if (example_df['cusip'][x] == example_df['cusip'][x-1]):

          # If bond matures, don't report book value
          if(example_df['par_val'][x] == 0):
          example_df['book_val'][x] = 0
          else:
          example_df['book_val'][x] = example_df['book_val'][x-1] - example_df['amortization'][x-1] - example_df['prin_cf'][x-1]


          example_df['amortization'][x] = (example_df['book_val'][x] - example_df['par_val'][x]) / example_df['wal'][x] / 12
          return example_df


          Numba:



          from numba import njit

          @njit
          def calculator(cusip, par, book, amort, prin_cf, wal):
          n = len(par)
          for i in range(1, n):
          if cusip[i] == cusip[i-1]:
          if par[i] == 0:
          book[i] == 0
          else:
          book[i] = book[i-1] - amort[i-1] - prin_cf[i-1]
          amort[i] = (book[i] - par[i]) / wal[i] / 12
          return book, amort


          def jpp(df):
          df['book_val'], df['amortization'] = calculator(pd.factorize(df['cusip'])[0], df['par_val'].values,
          df['book_val'].values, df['amortization'].values,
          df['prin_cf'].values, df['wal'].values)

          return df






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 13 '18 at 12:11









          jppjpp

          100k2161111




          100k2161111












          • Thanks! Although unfortunately numba is not on the list of libraries I have access to at work... any ideas on how the pandas/groupby solution might look?

            – M.Hope
            Dec 10 '18 at 15:29











          • @M.Hope, Sorry, can't think of anything. Get your work to install numba :)

            – jpp
            Dec 10 '18 at 15:35


















          • Thanks! Although unfortunately numba is not on the list of libraries I have access to at work... any ideas on how the pandas/groupby solution might look?

            – M.Hope
            Dec 10 '18 at 15:29











          • @M.Hope, Sorry, can't think of anything. Get your work to install numba :)

            – jpp
            Dec 10 '18 at 15:35

















          Thanks! Although unfortunately numba is not on the list of libraries I have access to at work... any ideas on how the pandas/groupby solution might look?

          – M.Hope
          Dec 10 '18 at 15:29





          Thanks! Although unfortunately numba is not on the list of libraries I have access to at work... any ideas on how the pandas/groupby solution might look?

          – M.Hope
          Dec 10 '18 at 15:29













          @M.Hope, Sorry, can't think of anything. Get your work to install numba :)

          – jpp
          Dec 10 '18 at 15:35






          @M.Hope, Sorry, can't think of anything. Get your work to install numba :)

          – jpp
          Dec 10 '18 at 15:35




















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