将标量值分配给空的 DataFrame 似乎没有任何作用 [英] Assigning a scalar value to an empty DataFrame doesn't appear to do anything

问题描述

我是熊猫新手，有一个非常基本的问题，拜托！

I'm new to pandas and have a very basic question, please!

通过 spyder 在 Python v3.6 上:

On Python v3.6 through spyder:

x= pd.DataFrame(columns = ['1','2'])
print(x)
x['1'] = '25'
print(x)

从打印语句来看，数据框 x 似乎没有改变.我的问题:x['1'] = '25' 有什么作用?

From the print statements, the dataframe x does not appear to change. My question: What does x['1'] = '25' do, if anything?

推荐答案

分配标量和可迭代对象的语义实际上是有区别的(将列表等容器视为类列表对象).

There is actually a difference between the semantics of assigning scalars and iterables (think containers such as lists as list-like objects).

考虑，

df = pd.DataFrame(columns=['1', '2'])                                                                                             
df                                                                                                                                  

Empty DataFrame
Columns: [1, 2]
Index: []

您已经定义了一个 空 数据框，没有任何索引(没有行)，但只有列的架构.

You've defined an empty dataframe without any index (no rows), but only a schema for the columns.

当您为列分配标量时，分配将在所有行中广播.在这种情况下，因为没有，所以什么也没有发生:

When you assign a scalar to a column, the assignment is broadcast across all rows. In this case, since there are none, nothing happens:

df['1'] = 123
df

Empty DataFrame
Columns: [1, 2]
Index: []

然而，分配一个类似列表的可迭代对象是另一回事，因为 Pandas 会为其创建新行:

However, assigning a list-like iterable is a different story, as pandas will create new rows for it:

df['1'] = [123]
df

     1    2
0  123  NaN

<小时>

现在，要了解标量分配的工作原理，请考虑一个类似的空 DataFrame，但具有已定义的索引:

---

Now, to understand how scalar assignment works, consider a similar empty DataFrame, but with a defined index:

df = pd.DataFrame(columns=['1', '2'], index=[0, 1])
df                                                                                                                                  

     1    2
0  NaN  NaN
1  NaN  NaN

它仍然是空的"(不是真的)，但现在我们可以分配标量并广播分配，

it is still "empty" (not really), but now we can assign scalars and the assignment is broadcast,

df['1'] = 123
df                                                                                                                                  

     1    2
0  123  NaN
1  123  NaN

将此行为与之前显示的行为进行对比.

Contrast this behaviour with that previously shown.

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