I have a column, 'col2', that has a list of strings. The current code I have is too slow, there's about 2000 unique strings (the letters in the example below), and 4000 rows. Ending up as 2000 columns and 4000 rows.
In [268]: df.head()
Out[268]:
col1 col2
0 6 A,B
1 15 C,G,A
2 25 B
Is there a fast way to make this in a get dummies format? Where each string has it's own column and in each string's column there is a 0 or 1 if it that row has that string in col2.
In [268]: def get_list(df):
d = []
for row in df.col2:
row_list = row.split(',')
for string in row_list:
if string not in d:
d.append(string)
return d
df_list = get_list(df)
def make_cols(df, lst):
for string in lst:
df[string] = 0
return df
df = make_cols(df, df_list)
for idx in range(0, len(df['col2'])):
row_list = df['col2'].iloc[idx].split(',')
for string in row_list:
df[string].iloc[idx]+= 1
Out[113]:
col1 col2 A B C G
0 6 A,B 1 1 0 0
1 15 C,G,A 1 0 1 1
2 25 B 0 1 0 0
This is my current code for it but it's too slow.
Thanks you any help!
You can use:
>>> df['col2'].str.get_dummies(sep=',')
A B C G
0 1 1 0 0
1 1 0 1 1
2 0 1 0 0
To join the Dataframes:
>>> pd.concat([df, df['col2'].str.get_dummies(sep=',')], axis=1)
col1 col2 A B C G
0 6 A,B 1 1 0 0
1 15 C,G,A 1 0 1 1
2 25 B 0 1 0 0
来源:https://stackoverflow.com/questions/28121682/quickest-way-to-make-a-get-dummies-type-dataframe-from-a-column-with-a-multiple