Split list into column

Question

I have a df like

uid                                  services
000c80b7d2b3643689b1e516918ec193    ['A']
001b292c588ec6cc11f57324d40e422d    ['B','A',C']
006696f65899fdd87ba4894c784716f9    ['C','B']

(not ordered list in services column)

I'd like to remap the list in columns

uid                                  services      A   B  C 
000c80b7d2b3643689b1e516918ec193    ['A']          1   0  0
001b292c588ec6cc11f57324d40e422d    ['B','A',C']   1   1  1
006696f65899fdd87ba4894c784716f9    ['C','B']      0   1  1

Thank you

Answer 1

import pandas as pd
d = pd.read_csv('TestingAccommodations.csv', encoding = 'ANSI')
q = dict(tuple(d.groupby('student_id')))

#%%

t1 =[]
t2 = []
t3 =[]
t4 =[]
t5 =[]
t6= []
t11 =[]
t12 =[]
t13 =[]
t14 = []
for k, v in q.items():
    t1.append(k)
    t2.append(v['last_nam'].iloc[0])
    t3.append(v['first_nam'].iloc[0])
    t4.append(v['school_dbn'].iloc[0])
    t5.append(v['official_class'].iloc[0])
    t6.append(v['grade_level'].iloc[0])
    t7 = []
    t8 = []
    t9 =[]
    t10 = []
    for i in v['Accommodation']:
        t7.append(i)
    for j in v['Description']:
        t8.append(j)
    for k in v['SpecificImpRecoms']:
        t9.append(k)
    for l in v['OtherAccommodation']:
        t10.append(l)
    t11.append(t7)
    t12.append(t8)
    t13.append(t9)
    t14.append(t10)
cols = d.columns.to_list()
new = pd.DataFrame({cols[0]:t1, cols[1]:t2,cols[2]:t3,cols[3]:t4,cols[4]:t5,cols[5]:t6,cols[6]:t11,
                    cols[7]:t12,cols[8]:t13,cols[9]:t14})
new.to_excel('new.xlsx',index = False)

#%%

new.columns
f = lambda x: 'Accomodation_{}'.format(x + 1)
op = pd.DataFrame(
    new.Accommodation.values.tolist(),
    new.index, dtype=str).fillna('').rename(columns=f)
f2 = lambda x: 'Description_{}'.format(x + 1)
op2 = pd.DataFrame(
    new.Description.values.tolist(),
    new.index, dtype=str).fillna('').rename(columns=f)
f3 = lambda x: 'SpecificImpRecoms_{}'.format(x + 1)
op3 = pd.DataFrame(
    new.SpecificImpRecoms.values.tolist(),
    new.index, dtype=str).fillna('').rename(columns=f)
f4 = lambda x: 'OtherAccommodation_{}'.format(x + 1)
op4 = pd.DataFrame(
    new.OtherAccommodation.values.tolist(),
    new.index, dtype=str).fillna('').rename(columns=f)

new2 = pd.concat([new,op,op2,op3,op4], axis = 1)

Answer 2

Quick Answer:

df.join(df['services'].apply(lambda x: "|".join(x)).str.get_dummies())

---

One way is to turn the list of characters into a delimited string (using the pipe symbol | here) and use pd.Series.str.get_dummies:

df = pd.DataFrame([[['A']],[list('ABC')],[list('BC')]],
                  columns=['services'],
                  index=['abc','def','ghi'])
df.index.name = 'UID'
df

      services
UID           
abc        [A]
def  [A, B, C]
ghi     [B, C]

(df['services']
 .apply(lambda x: "|".join(x))
 .str.get_dummies())

     A  B  C
UID         
abc  1  0  0
def  1  1  1
ghi  0  1  1

Merging into the original then becomes a one-liner:

df.join(df['services'].apply(lambda x: "|".join(x)).str.get_dummies())
      services  A  B  C
UID                    
abc        [A]  1  0  0
def  [A, B, C]  1  1  1
ghi     [B, C]  0  1  1

Answer 3

You can use MultiLabelBinarizer first and then join:

from sklearn.preprocessing import MultiLabelBinarizer
mlb = MultiLabelBinarizer()

print (pd.DataFrame(mlb.fit_transform(df['services']),columns=mlb.classes_, index=df.index))
   A  B  C
0  1  0  0
1  1  1  1
2  0  1  1

df1 = pd.DataFrame(mlb.fit_transform(df['services']),columns=mlb.classes_, index=df.index)
df = df.join(df1)
print (df)
                                uid   services  A  B  C
0  000c80b7d2b3643689b1e516918ec193        [A]  1  0  0
1  001b292c588ec6cc11f57324d40e422d  [B, A, C]  1  1  1
2  006696f65899fdd87ba4894c784716f9     [C, B]  0  1  1

Pure pandas alternative with get_dummies and groupby by columns with aggregate max:

df1 = pd.get_dummies(pd.DataFrame(df['services'].values.tolist()), prefix='', prefix_sep='')
        .groupby(axis=1, level=0).max()
print (df1)
   A  B  C
0  1  0  0
1  1  1  1
2  0  1  1

df = df.join(df1)
print (df)
                                uid   services  A  B  C
0  000c80b7d2b3643689b1e516918ec193        [A]  1  0  0
1  001b292c588ec6cc11f57324d40e422d  [B, A, C]  1  1  1
2  006696f65899fdd87ba4894c784716f9     [C, B]  0  1  1

Timings:

#3k rows 
df = pd.concat([df]*1000).reset_index(drop=True)

#John Galt solution
In [255]: %timeit (df.join(df.services.apply(lambda x: pd.Series({y:1 for y in x})).fillna(0).astype(int)))
1 loop, best of 3: 658 ms per loop

#user1717828 solution
In [256]: %timeit (df.join(df['services'].apply(lambda x: "|".join(x)).str.get_dummies()))
100 loops, best of 3: 16.8 ms per loop

#Jez solution1
In [257]: %timeit (df.join(pd.DataFrame(mlb.fit_transform(df['services']),columns=mlb.classes_, index=df.index)))
100 loops, best of 3: 4.66 ms per loop

#Jez solution2
In [258]: %timeit (df.join(pd.get_dummies(pd.DataFrame(df['services'].values.tolist()), prefix='', prefix_sep='').groupby(axis=1, level=0).max()))
100 loops, best of 3: 7.04 ms per loop

---

#30k rows
df = pd.concat([df]*10000).reset_index(drop=True)


#John Galt solution
In [260]: %timeit (df.join(df.services.apply(lambda x: pd.Series({y:1 for y in x})).fillna(0).astype(int)))
1 loop, best of 3: 6.68 s per loop

#user1717828 solution
In [261]: %timeit (df.join(df['services'].apply(lambda x: "|".join(x)).str.get_dummies()))
10 loops, best of 3: 138 ms per loop

#Jez solution1
In [262]: %timeit (df.join(pd.DataFrame(mlb.fit_transform(df['services']),columns=mlb.classes_, index=df.index)))
10 loops, best of 3: 39.8 ms per loop

#Jez solution2
In [263]: %timeit (df.join(pd.get_dummies(pd.DataFrame(df['services'].values.tolist()), prefix='', prefix_sep='').groupby(axis=1, level=0).max()))
10 loops, best of 3: 20.6 ms per loop

Answer 4

df['A'] = list(map(lambda x: 1 if 'A' in x else 0, df['Services'].tolist()))

df['B'] = list(map(lambda x: 1 if 'B' in x else 0, df['Services'].tolist()))

df['C'] = list(map(lambda x: 1 if 'C' in x else 0, df['Services'].tolist()))

Answer 5

In [1158]: df.join(df.services.apply(lambda x: pd.Series({y:1 for y in x})).fillna(0))
Out[1158]:
                                uid   services    A    B    C
0  000c80b7d2b3643689b1e516918ec193        [A]  1.0  0.0  0.0
1  001b292c588ec6cc11f57324d40e422d  [B, A, C]  1.0  1.0  1.0
2  006696f65899fdd87ba4894c784716f9     [C, B]  0.0  1.0  1.0