I have columns in a dataframe (imported from a CSV) containing text like this.
"New york", "Atlanta", "Mumbai"
"Beijing", "Paris", "Budapest"
"Brussels", "Oslo", "Singapore"
I want to collapse/merge all the columns into one single column, like this
New york Atlanta
Beijing Paris Budapest
Brussels Oslo Singapore
How to do it in pandas?
A faster (but uglier) version is with .cat:
df[0].str.cat(df.ix[:, 1:].T.values, sep=' ')
0 New york Atlanta Mumbai
1 Beijing Paris Budapest
2 Brussels Oslo Singapore
Name: 0, dtype: object
On a larger (10kx5) DataFrame:
%timeit df.apply(" ".join, axis=1)
10 loops, best of 3: 112 ms per loop
%timeit df[0].str.cat(df.ix[:, 1:].T.values, sep=' ')
100 loops, best of 3: 4.48 ms per loop
Suppose you have a DataFrame like so:
>>> df
0 1 2
0 New york Atlanta Mumbai
1 Beijing Paris Budapest
2 Brussels Oslo Singapore
Then, a simple use of the pd.DataFrame.apply method will work nicely:
>>> df.apply(" ".join, axis=1)
0 New york Atlanta Mumbai
1 Beijing Paris Budapest
2 Brussels Oslo Singapore
dtype: object
Note, I have to pass axis=1 so that it is applied across the columns, rather than down the rows. I.e:
>>> df.apply(" ".join, axis=0)
0 New york Beijing Brussels
1 Atlanta Paris Oslo
2 Mumbai Budapest Singapore
dtype: object
df.T.apply(" ".join)Here are a couple more ways:
def pir(df):
df = df.copy()
df.insert(2, 's', ' ', 1)
df.insert(1, 's', ' ', 1)
return df.sum(1)
def pir2(df):
df = df.copy()
return pd.MultiIndex.from_arrays(df.values.T).to_series().str.join(' ').reset_index(drop=True)
def pir3(df):
a = df.values[:, 0].copy()
for j in range(1, df.shape[1]):
a += ' ' + df.values[:, j]
return pd.Series(a)
pir3 seems fastest over small df
pir3 still fastest over larger df 30,000 rows
pir3 should be concatenating df.values[ : , 0], df.values[ : , 1] ...pd.concat? or np.concatenate? Neither of those combine strings. I'd have to use join. Unless I misunderstood what you meant.+ operator. You fixed it now anyway.If you prefer something more explicit...
Starting with a dataframe df that looks like this:
>>> df
A B C
0 New york Beijing Brussels
1 Atlanta Paris Oslo
2 Mumbai Budapest Singapore
You can create a new column like this:
df['result'] = df['A'] + ' ' + df['B'] + ' ' + df['C']
In this case the result is stored in the 'result' column of the original DataFrame:
A B C result
0 New york Beijing Brussels New york Beijing Brussels
1 Atlanta Paris Oslo Atlanta Paris Oslo
2 Mumbai Budapest Singapore Mumbai Budapest Singapore
for the sake of completeness:
In [160]: df1.add([' '] * (df1.columns.size - 1) + ['']).sum(axis=1)
Out[160]:
0 New york Atlanta Mumbai
1 Beijing Paris Budapest
2 Brussels Oslo Singapore
dtype: object
Explanation:
In [162]: [' '] * (df.columns.size - 1) + ['']
Out[162]: [' ', ' ', '']
Timing against 300K rows DF:
In [68]: df = pd.concat([df] * 10**5, ignore_index=True)
In [69]: df.shape
Out[69]: (300000, 3)
In [76]: %timeit df.apply(" ".join, axis=1)
1 loop, best of 3: 5.8 s per loop
In [77]: %timeit df[0].str.cat(df.ix[:, 1:].T.values, sep=' ')
10 loops, best of 3: 138 ms per loop
In [79]: %timeit pir(df)
1 loop, best of 3: 499 ms per loop
In [80]: %timeit pir2(df)
10 loops, best of 3: 174 ms per loop
In [81]: %timeit pir3(df)
10 loops, best of 3: 115 ms per loop
In [159]: %timeit df.add([' '] * (df.columns.size - 1) + ['']).sum(axis=1)
1 loop, best of 3: 478 ms per loop
Conclusion: current winner is @piRSquared's pir3()
