I have a dataframe similar to this:
data = {"col_1": [0, 1, 2],
"col_2": ["abc", "defg", "hi"]}
df = pd.DataFrame(data)
Visually:
col_1 col_2
0 0 abc
1 1 defg
2 2 hi
What I'd like to do is split up each character in col_2, and append it as a new column to the dataframe
example iterative method:
def get_chars(string):
chars = []
for char in string:
chars.append(char)
return chars
char_df = pd.DataFrame()
for i in range(len(df)):
char_arr = get_chars(df.loc[i, "col_2"])
temp_df = pd.DataFrame(char_arr).T
char_df = pd.concat([char_df, temp_df], ignore_index=True, axis=0)
df = pd.concat([df, char_df], ignore_index=True, axis=1)
Which results in the correct form:
0 1 2 3 4 5
0 0 abc a b c NaN
1 1 defg d e f g
2 2 hi h i NaN NaN
But I believe iterating though the dataframe like this is very inefficient, so I want to find a faster (ideally vectorised) solution.
In reality, I'm not really splitting up strings, but the point of this question is to find a way to efficiently process one column, and return many.
If need performance use DataFrame constructor with convert values to lists:
df = df.join(pd.DataFrame([list(x) for x in df['col_2']], index=df.index))
Or:
df = df.join(pd.DataFrame(df['col_2'].apply(list).tolist(), index=df.index))
print (df)
col_1 col_2 0 1 2 3
0 0 abc a b c None
1 1 defg d e f g
2 2 hi h i None None
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