How to identify array records that has value in dataframe

Question

my dataframe df looks like this

       RepID   Arr
0      3       [E83, J99]
1      5       [P81, G39, E34]
2      6       [R19]
...
...
1857   8885    [G39, F14, T22]

I want to find all records that have the value "G39" in the array

       RepID   Arr
1      5       [P81, G39, E34]
1857   8885    [G39, F14, T22]

I tried this but it did not work

np.where(df["Arr"]=="G39")

I get empty

(array([], dtype=int64),)

How to get that list?

Asked By: asmgx

||

Source

Answer 1

You can use list comprehension to iterate through array column and filter and collect dataframe records based on condition:

import pandas as pd

data = [
  (3, ["E83", "J99"]),
  (5, ["P81", "G39", "E34"]),
  (6, ["R19"]),
  (8885, ["G39", "F14", "T22"]),
]

df = pd.DataFrame(data=data, columns=["RepID", "Arr"])

df[[("G39" in typ_arr) for typ_arr in df["Arr"]]]
>>     RepID               Arr
>> 1       5   [P81, G39, E34]
>> 3    8885   [G39, F14, T22]

Answered By: Azhar Khan

Answer 2

The code below creates both, a dictionary and a pandas data frame with the requested records:

import pandas as pd
from pprint import pprint
df_print="""
       RepID   Arr
0      3       [E83, J99]
1      5       [P81, G39, E34]
2      6       [R19]
3   8885       [G39, F14, T22]"""
open('df_print', 'w').write(df_print)
df = pd.read_table('df_print', sep=r'sss*', index_col = 0)

dct_rows = {}
for index, row in df.iterrows():
   if "G39" in row["Arr"]:
       dct_rows[index] = { 'RepID' : row['RepID'], "Arr" : row["Arr"] }
print("dct_rows = \n", end='')
pprint(dct_rows)
print("---")
df_G39 = pd.DataFrame.from_dict(dct_rows, orient='index')
print("df_G39_print = """\")
print(df_G39, end='"""n')

printing:

dct_rows = 
{1: {'Arr': '[P81, G39, E34]', 'RepID': 5},
 3: {'Arr': '[G39, F14, T22]', 'RepID': 8885}}
---
df_G39_print = """
   RepID              Arr
1      5  [P81, G39, E34]
3   8885  [G39, F14, T22]"""

Answered By: Claudio

How to identify array records that has value in dataframe

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