Question

我有这个df：

l1 = ['a', 'b', 'c']
l2 = ['x', ['y1', 'y2', 'y3'], 'z']
df = pd.DataFrame(list(zip(l1, l2)), columns = ['l1', 'l2'])

结果：

  l1            l2
0  a             x
1  b  [y1, y2, y3]
2  c             z

我需要的是在l2中解压缩内部列表并在l1中传播相应的值，如下所示：

  l1  l2
0  a   x
1  b  y1
2  b  y2
3  b  y3
4  c   z

这样做的正确方法是什么？感谢。

Answer 1

您可以将嵌套列表理解与itertools.zip_longest一起使用。

import pandas as pd

from itertools import zip_longest

l1 = ['a', 'b', 'c']
l2 = ['x', ['y1', 'y2', 'y3'], 'z']

expanded = [(left, right) for outer in zip(l1, l2) 
                          for left, right in zip_longest(*outer, fillvalue=outer[0])]

pd.DataFrame(expanded)

结果是......

对我来说，这是一个列表组合太长的边界。还假设l1中没有列表，并且正在填写。

Answer 2

蛮力，循环遍历数据框：

for idx in df.index:
    # This transforms the item in "l2" into an iterable list
    item = df.loc[idx, "l2"] if isinstance(df.loc[idx, "l2"], (list, tuple)) else [df.loc[idx, "l2"]]
    for element in item:
        print(df.loc[idx, "l1"], element)

返回

a x
b y1
b y2
b y3
c z

Answer 3

我认为您可以numpy.repeat使用str.len来表示重复值，lists可以使用嵌套chain的平面值：

from  itertools import chain

df1 = pd.DataFrame({
        "l1": np.repeat(df.l1.values, df.l2.str.len()),
        "l2": list(chain.from_iterable(df.l2))})
print (df1)
  l1  l2
0  a   x
1  b  y1
2  b  y2
3  b  y3
4  c   z

<强>计时：

#[100000 rows x 2 columns]
np.random.seed(10)
N = 100000
l1 = ['a', 'b', 'c']
l1 = np.random.choice(l1, N)
l2 = [list(tuple(string.ascii_letters[:np.random.randint(1, 10)])) for _ in np.arange(N)]
df = pd.DataFrame({"l1":l1, "l2":l2})
df.l2 = df.l2.apply(lambda x: x if len(x) !=1 else x[0])
#print (df)


In [91]: %timeit (pd.DataFrame([(left, right) for outer in zip(l1, l2) for left, right in zip_longest(*outer, fillvalue=outer[0])]))
1 loop, best of 3: 242 ms per loop

In [92]: %timeit (pd.DataFrame({ "l1": np.repeat(df.l1.values, df.l2.str.len()), "l2": list(chain.from_iterable(df.l2))}))
10 loops, best of 3: 84.6 ms per loop

<强>结论：

numpy.repeat 3 times更快zip_longest解决更大的df。

编辑：

为了与循环版本进行比较，需要更小的df，因为非常慢：

#[1000 rows x 2 columns]
np.random.seed(10)
N = 1000
l1 = ['a', 'b', 'c']
l1 = np.random.choice(l1, N)
l2 = [list(tuple(string.ascii_letters[:np.random.randint(1, 10)])) for _ in np.arange(N)]
df = pd.DataFrame({"l1":l1, "l2":l2})
df.l2 = df.l2.apply(lambda x: x if len(x) !=1 else x[0])
#print (df)

def alexey(df):
    df2 = pd.DataFrame(columns=df.columns,index=df.index)[0:0]

    for idx in df.index:
        new_row = df.loc[idx, :].copy()
        for res in df.ix[idx, 'l2']:
            new_row.set_value('l2', res)
            df2.loc[len(df2)] = new_row
    return df2

print (alexey(df))

In [20]: %timeit (alexey(df))
1 loop, best of 3: 11.4 s per loop

In [21]: %timeit pd.DataFrame([(left, right) for outer in zip(l1, l2) for left, right in zip_longest(*outer, fillvalue=outer[0])])
100 loops, best of 3: 2.57 ms per loop

In [22]: %timeit pd.DataFrame({ "l1": np.repeat(df.l1.values, df.l2.str.len()), "l2": list(chain.from_iterable(df.l2))})
The slowest run took 4.42 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 1.41 ms per loop

Answer 4

对于具有不恒定列数的DataFrame，我现在执行以下操作：

l1 = ['a', 'b', 'c']
l2 = ['x', ['y1', 'y2', 'y3'], 'z']
df = pd.DataFrame(list(zip(l1, l2)), columns = ['l1', 'l2'])

df2 = pd.DataFrame(columns=df.columns,index=df.index)[0:0]

for idx in df.index:
    new_row = df.loc[idx, :].copy()
    for res in df.ix[idx, 'l2']:
        new_row.set_value('l2', res)
        df2.loc[len(df2)] = new_row

它有效，但这看起来非常像暴力。

解压缩DataFrame的list元素

4 个答案: