pandas 将空列表列添加到 DataFrame
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Add column of empty lists to DataFrame
提问by vk1011
Similar to this question How to add an empty column to a dataframe?, I am interested in knowing the best way to add a column of empty lists to a DataFrame.
类似于这个问题如何将空列添加到数据框？，我有兴趣了解将一列空列表添加到 DataFrame 的最佳方法。
What I am trying to do is basically initialize a column and as I iterate over the rows to process some of them, then add a filled list in this new column to replace the initialized value.
我想要做的基本上是初始化一个列，当我遍历行以处理其中的一些时，然后在这个新列中添加一个填充列表来替换初始化值。
For example, if below is my initial DataFrame:
例如，如果下面是我的初始 DataFrame：
df = pd.DataFrame(d = {'a': [1,2,3], 'b': [5,6,7]}) # Sample DataFrame
>>> df
a b
0 1 5
1 2 6
2 3 7
Then I want to ultimately end up with something like this, where each row has been processed separately (sample results shown):
然后我想最终得到这样的结果，其中每一行都被单独处理（显示了示例结果）：
>>> df
a b c
0 1 5 [5, 6]
1 2 6 [9, 0]
2 3 7 [1, 2, 3]
Of course, if I try to initialize like df['e'] = []as I would with any other constant, it thinks I am trying to add a sequence of items with length 0, and hence fails.
当然，如果我尝试像df['e'] = []使用任何其他常量一样初始化，它认为我正在尝试添加长度为 0 的项目序列，因此失败。
If I try initializing a new column as Noneor NaN, I run in to the following issues when trying to assign a list to a location.
如果我尝试将新列初始化为None或NaN，则在尝试将列表分配给某个位置时会遇到以下问题。
df['d'] = None
>>> df
a b d
0 1 5 None
1 2 6 None
2 3 7 None
Issue 1 (it would be perfect if I can get this approach to work! Maybe something trivial I am missing):
问题 1（如果我能用这种方法就完美了！也许我遗漏了一些微不足道的东西）：
>>> df.loc[0,'d'] = [1,3]
...
ValueError: Must have equal len keys and value when setting with an iterable
Issue 2 (this one works, but not without a warning because it is not guaranteed to work as intended):
问题 2（这个有效，但并非没有警告，因为不能保证按预期工作）：
>>> df['d'][0] = [1,3]
C:\Python27\Scripts\ipython:1: SettingWithCopyWarning:
A value is trying to be set on a copy of a slice from a DataFrame
Hence I resort to initializing with empty lists and extending them as needed. There are a couple of methods I can think of to initialize this way, but is there a more straightforward way?
因此，我使用空列表进行初始化并根据需要扩展它们。我可以想到几种方法来以这种方式初始化，但是有没有更直接的方法？
Method 1:
方法一：
df['empty_lists1'] = [list() for x in range(len(df.index))]
>>> df
a b empty_lists1
0 1 5 []
1 2 6 []
2 3 7 []
Method 2:
方法二：
df['empty_lists2'] = df.apply(lambda x: [], axis=1)
>>> df
a b empty_lists1 empty_lists2
0 1 5 [] []
1 2 6 [] []
2 3 7 [] []
Summary of questions:
问题总结：
Is there any minor syntax change that can be addressed in Issue 1 that can allow a list to be assigned to a None/NaNinitialized field?
是否有任何可以在问题 1 中解决的小的语法更改可以允许将列表分配给None/NaN初始化字段？
If not, then what is the best way to initialize a new column with empty lists?
如果不是，那么用空列表初始化新列的最佳方法是什么？
回答by ComputerFellow
One more way is to use np.empty:
另一种方法是使用np.empty：
df['empty_list'] = np.empty((len(df), 0)).tolist()
You could also knock off .indexin your "Method 1" when trying to find lenof df.
你也可以收工.index试图找到当你的“方法1”len的df。
df['empty_list'] = [[] for _ in range(len(df))]
Turns out, np.emptyis faster...
事实证明，np.empty速度更快......
In [1]: import pandas as pd
In [2]: df = pd.DataFrame(pd.np.random.rand(1000000, 5))
In [3]: timeit df['empty1'] = pd.np.empty((len(df), 0)).tolist()
10 loops, best of 3: 127 ms per loop
In [4]: timeit df['empty2'] = [[] for _ in range(len(df))]
10 loops, best of 3: 193 ms per loop
In [5]: timeit df['empty3'] = df.apply(lambda x: [], axis=1)
1 loops, best of 3: 5.89 s per loop
回答by tozCSS
I timed all the three methods in the accepted answer, the fastest one took 216 ms on my machine. However, this took only 28 ms:
我对接受的答案中的所有三种方法进行了计时，最快的一种在我的机器上花费了 216 毫秒。但是，这仅用了 28 毫秒：
df['empty4'] = [[]] * len(df)
df['empty4'] = [[]] * len(df)
Note: Similarly, df['e5'] = [set()] * len(df)also took 28ms.
注：同理，df['e5'] = [set()] * len(df)也用了28ms。
