如何将 Dataframe 单元格中的列表爆炸成单独的行

小开

I think this a really good question, in Hive you would use EXPLODE, I think there is a case to be made that Pandas should include this functionality by default. I would probably explode the list column with a nested generator comprehension like this:

pd.DataFrame({
"name": i[0],
"opponent": i[1],
"nearest_neighbor": neighbour
}
for i, row in df.iterrows() for neighbour in row.nearest_neighbors
).set_index(["name", "opponent"])

小开

最佳答案

In the code below, I first reset the index to make the row iteration easier.

I create a list of lists where each element of the outer list is a row of the target DataFrame and each element of the inner list is one of the columns. This nested list will ultimately be concatenated to create the desired DataFrame.

I use a lambda function together with list iteration to create a row for each element of the nearest_neighbors paired with the relevant name and opponent.

Finally, I create a new DataFrame from this list (using the original column names and setting the index back to name and opponent).

df = (pd.DataFrame({'name': ['A.J. Price'] * 3,
'opponent': ['76ers', 'blazers', 'bobcats'],
'nearest_neighbors': [['Zach LaVine', 'Jeremy Lin', 'Nate Robinson', 'Isaia']] * 3})
.set_index(['name', 'opponent']))


>>> df
nearest_neighbors
name       opponent
A.J. Price 76ers     [Zach LaVine, Jeremy Lin, Nate Robinson, Isaia]
blazers   [Zach LaVine, Jeremy Lin, Nate Robinson, Isaia]
bobcats   [Zach LaVine, Jeremy Lin, Nate Robinson, Isaia]


df.reset_index(inplace=True)
rows = []
_ = df.apply(lambda row: [rows.append([row['name'], row['opponent'], nn])
for nn in row.nearest_neighbors], axis=1)
df_new = pd.DataFrame(rows, columns=df.columns).set_index(['name', 'opponent'])


>>> df_new
nearest_neighbors
name       opponent
A.J. Price 76ers          Zach LaVine
76ers           Jeremy Lin
76ers        Nate Robinson
76ers                Isaia
blazers        Zach LaVine
blazers         Jeremy Lin
blazers      Nate Robinson
blazers              Isaia
bobcats        Zach LaVine
bobcats         Jeremy Lin
bobcats      Nate Robinson
bobcats              Isaia

EDIT JUNE 2017

An alternative method is as follows:

>>> (pd.melt(df.nearest_neighbors.apply(pd.Series).reset_index(),
id_vars=['name', 'opponent'],
value_name='nearest_neighbors')
.set_index(['name', 'opponent'])
.drop('variable', axis=1)
.dropna()
.sort_index()
)

小开

Nicer alternative solution with apply(pd.Series):

df = pd.DataFrame({'listcol':[[1,2,3],[4,5,6]]})


# expand df.listcol into its own dataframe
tags = df['listcol'].apply(pd.Series)


# rename each variable is listcol
tags = tags.rename(columns = lambda x : 'listcol_' + str(x))


# join the tags dataframe back to the original dataframe
df = pd.concat([df[:], tags[:]], axis=1)

小开

Similar to Hive's EXPLODE functionality:

import copy


def pandas_explode(df, column_to_explode):
"""
Similar to Hive's EXPLODE function, take a column with iterable elements, and flatten the iterable to one element
per observation in the output table


:param df: A dataframe to explod
:type df: pandas.DataFrame
:param column_to_explode:
:type column_to_explode: str
:return: An exploded data frame
:rtype: pandas.DataFrame
"""


# Create a list of new observations
new_observations = list()


# Iterate through existing observations
for row in df.to_dict(orient='records'):


# Take out the exploding iterable
explode_values = row[column_to_explode]
del row[column_to_explode]


# Create a new observation for every entry in the exploding iterable & add all of the other columns
for explode_value in explode_values:


# Deep copy existing observation
new_observation = copy.deepcopy(row)


# Add one (newly flattened) value from exploding iterable
new_observation[column_to_explode] = explode_value


# Add to the list of new observations
new_observations.append(new_observation)


# Create a DataFrame
return_df = pandas.DataFrame(new_observations)


# Return
return return_df

小开

Use apply(pd.Series) and stack, then reset_index and to_frame

In [1803]: (df.nearest_neighbors.apply(pd.Series)
.stack()
.reset_index(level=2, drop=True)
.to_frame('nearest_neighbors'))
Out[1803]:
nearest_neighbors
name       opponent
A.J. Price 76ers          Zach LaVine
76ers           Jeremy Lin
76ers        Nate Robinson
76ers                Isaia
blazers        Zach LaVine
blazers         Jeremy Lin
blazers      Nate Robinson
blazers              Isaia
bobcats        Zach LaVine
bobcats         Jeremy Lin
bobcats      Nate Robinson
bobcats              Isaia

Details

In [1804]: df
Out[1804]:
nearest_neighbors
name       opponent
A.J. Price 76ers     [Zach LaVine, Jeremy Lin, Nate Robinson, Isaia]
blazers   [Zach LaVine, Jeremy Lin, Nate Robinson, Isaia]
bobcats   [Zach LaVine, Jeremy Lin, Nate Robinson, Isaia]

小开

Here is a potential optimization for larger dataframes. This runs faster when there are several equal values in the "exploding" field. (The larger the dataframe is compared to the unique value count in the field, the better this code will perform.)

def lateral_explode(dataframe, fieldname):
temp_fieldname = fieldname + '_made_tuple_'
dataframe[temp_fieldname] = dataframe[fieldname].apply(tuple)
list_of_dataframes = []
for values in dataframe[temp_fieldname].unique().tolist():
list_of_dataframes.append(pd.DataFrame({
temp_fieldname: [values] * len(values),
fieldname: list(values),
}))
dataframe = dataframe[list(set(dataframe.columns) - set([fieldname]))]\
.merge(pd.concat(list_of_dataframes), how='left', on=temp_fieldname)
del dataframe[temp_fieldname]


return dataframe

小开

The fastest method I found so far is extending the DataFrame with .iloc and assigning back the flattened target column.

Given the usual input (replicated a bit):

df = (pd.DataFrame({'name': ['A.J. Price'] * 3,
'opponent': ['76ers', 'blazers', 'bobcats'],
'nearest_neighbors': [['Zach LaVine', 'Jeremy Lin', 'Nate Robinson', 'Isaia']] * 3})
.set_index(['name', 'opponent']))
df = pd.concat([df]*10)


df
Out[3]:
nearest_neighbors
name       opponent
A.J. Price 76ers     [Zach LaVine, Jeremy Lin, Nate Robinson, Isaia]
blazers   [Zach LaVine, Jeremy Lin, Nate Robinson, Isaia]
bobcats   [Zach LaVine, Jeremy Lin, Nate Robinson, Isaia]
76ers     [Zach LaVine, Jeremy Lin, Nate Robinson, Isaia]
blazers   [Zach LaVine, Jeremy Lin, Nate Robinson, Isaia]
...

Given the following suggested alternatives:

col_target = 'nearest_neighbors'


def extend_iloc():
# Flatten columns of lists
col_flat = [item for sublist in df[col_target] for item in sublist]
# Row numbers to repeat
lens = df[col_target].apply(len)
vals = range(df.shape[0])
ilocations = np.repeat(vals, lens)
# Replicate rows and add flattened column of lists
cols = [i for i,c in enumerate(df.columns) if c != col_target]
new_df = df.iloc[ilocations, cols].copy()
new_df[col_target] = col_flat
return new_df


def melt():
return (pd.melt(df[col_target].apply(pd.Series).reset_index(),
id_vars=['name', 'opponent'],
value_name=col_target)
.set_index(['name', 'opponent'])
.drop('variable', axis=1)
.dropna()
.sort_index())


def stack_unstack():
return (df[col_target].apply(pd.Series)
.stack()
.reset_index(level=2, drop=True)
.to_frame(col_target))

I find that extend_iloc() is the fastest:

%timeit extend_iloc()
3.11 ms ± 544 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)


%timeit melt()
22.5 ms ± 1.25 ms per loop (mean ± std. dev. of 7 runs, 100 loops each)


%timeit stack_unstack()
11.5 ms ± 410 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

小开

Extending Oleg's .iloc answer to automatically flatten all list-columns:

def extend_iloc(df):
cols_to_flatten = [colname for colname in df.columns if
isinstance(df.iloc[0][colname], list)]
# Row numbers to repeat
lens = df[cols_to_flatten[0]].apply(len)
vals = range(df.shape[0])
ilocations = np.repeat(vals, lens)
# Replicate rows and add flattened column of lists
with_idxs = [(i, c) for (i, c) in enumerate(df.columns) if c not in cols_to_flatten]
col_idxs = list(zip(*with_idxs)[0])
new_df = df.iloc[ilocations, col_idxs].copy()


# Flatten columns of lists
for col_target in cols_to_flatten:
col_flat = [item for sublist in df[col_target] for item in sublist]
new_df[col_target] = col_flat


return new_df

This assumes that each list-column has equal list length.

小开

So all of these answers are good but I wanted something ^really simple^ so here's my contribution:

def explode(series):
return pd.Series([x for inner_list in series for x in inner_list])

That's it.. just use this when you want a new series where the lists are 'exploded'. Here's an example where we do value_counts()

In[1]: df = pd.DataFrame({'column': [['a','b','c'],['b','c'],['c']]})
In [2]: df
Out[2]:
column
0  [a, b, c]
1     [b, c]
2        [c]


In [3]: explode(df['column'])
Out[3]:
0    a
1    b
2    c
3    b
4    c
5    c


In [4]: explode(df['column']).value_counts()
Out[4]:
c    3
b    2
a    1

小开

Exploding a list-like column has been simplified significantly in pandas 0.25 with the addition of the explode() method:

df = (pd.DataFrame({'name': ['A.J. Price'] * 3,
'opponent': ['76ers', 'blazers', 'bobcats'],
'nearest_neighbors': [['Zach LaVine', 'Jeremy Lin', 'Nate Robinson', 'Isaia']] * 3})
.set_index(['name', 'opponent']))


df.explode('nearest_neighbors')

Out:

                    nearest_neighbors
name       opponent
A.J. Price 76ers          Zach LaVine
76ers           Jeremy Lin
76ers        Nate Robinson
76ers                Isaia
blazers        Zach LaVine
blazers         Jeremy Lin
blazers      Nate Robinson
blazers              Isaia
bobcats        Zach LaVine
bobcats         Jeremy Lin
bobcats      Nate Robinson
bobcats              Isaia

小开

Instead of using apply(pd.Series) you can flatten the column. This improves performance.

df = (pd.DataFrame({'name': ['A.J. Price'] * 3,
'opponent': ['76ers', 'blazers', 'bobcats'],
'nearest_neighbors': [['Zach LaVine', 'Jeremy Lin', 'Nate Robinson', 'Isaia']] * 3})
.set_index(['name', 'opponent']))






%timeit (pd.DataFrame(df['nearest_neighbors'].values.tolist(), index = df.index)
.stack()
.reset_index(level = 2, drop=True).to_frame('nearest_neighbors'))


1.87 ms ± 9.74 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)




%timeit (df.nearest_neighbors.apply(pd.Series)
.stack()
.reset_index(level=2, drop=True)
.to_frame('nearest_neighbors'))


2.73 ms ± 16.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

小开

Panda's data frame have method explode for this.

df = (pd.DataFrame({'name': ['A.J. Price'] * 3,
'opponent': ['76ers', 'blazers', 'bobcats'],
'nearest_neighbors': [['Zach LaVine', 'Jeremy Lin', 'Nate Robinson', 'Isaia']] * 3}))


df.explode('nearest_neighbors')