将空列表列添加到DataFrame [英] Add column of empty lists to DataFrame
问题描述
类似于此问题如何向其中添加空列一个数据框?，我想知道向DataFrame添加一列空列表的最佳方法.
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.
我要做的基本上是初始化一列，并在行上进行迭代以处理其中的一些行，然后在此新列中添加填充列表以替换初始化的值.
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.
例如，如果下面是我的初始DataFrame:
For example, if below is my initial 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
然后我想最终得到这样的结果，其中每一行都经过单独处理(显示了示例结果):
Then I want to ultimately end up with something like this, where each row has been processed separately (sample results shown):
>>> df
a b c
0 1 5 [5, 6]
1 2 6 [9, 0]
2 3 7 [1, 2, 3]
当然，如果我尝试像df['e'] = []那样使用任何其他常量进行初始化，它会认为我正在尝试添加长度为0的项目序列，因此失败.
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.
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位置时遇到以下问题.
If I try initializing a new column as None or NaN, I run in to the following issues when trying to assign a list to a location.
df['d'] = None
>>> df
a b d
0 1 5 None
1 2 6 None
2 3 7 None
问题1(如果我可以采用这种方法，那将是完美的！也许我没想到一些琐碎的事情):
Issue 1 (it would be perfect if I can get this approach to work! Maybe something trivial I am missing):
>>> df.loc[0,'d'] = [1,3]
...
ValueError: Must have equal len keys and value when setting with an iterable
问题2(该方法有效，但并非没有警告，因为不能保证它可以按预期工作):
Issue 2 (this one works, but not without a warning because it is not guaranteed to work as intended):
>>> 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
因此，我诉诸于用空列表初始化并根据需要扩展它们.我可以想到几种方法来进行这种初始化，但是还有更简单的方法吗?
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?
方法1:
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 []
方法2:
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 [] []
问题摘要:
在问题1中是否可以解决任何小的语法更改，从而可以将列表分配给None/NaN初始化字段?
Is there any minor syntax change that can be addressed in Issue 1 that can allow a list to be assigned to a None/NaN initialized field?
如果没有，那么用空列表初始化新列的最佳方法是什么?
If not, then what is the best way to initialize a new column with empty lists?
推荐答案
One more way is to use np.empty:
df['empty_list'] = np.empty((len(df), 0)).tolist()
当您尝试找到df的len时，也可以在方法1"中关闭.index.
You could also knock off .index in your "Method 1" when trying to find len of df.
df['empty_list'] = [[] for _ in range(len(df))]
结果是np.empty更快...
Turns out, np.empty is faster...
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
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