blocks|key|1681173|text|当您使用apply时，它将per_country['type']的每个记录隐式地传递给func。如果您想要更简单、更清晰地这样做，可以使用lambda函数。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1681174|per_country['virus_confirmed']+=+per_country.apply(lambda+x:+x['cases']+if+x['type']=="confirmed"+else+0,+axis=1)|code-block|syntax|javascript|1681175|编辑:对于应用函数，请注意，您将它应用于DataFrame中的一个系列。这意味着x是来自该列的每个记录，您不需要指定轴。我将其应用于整个DataFrame，这意味着我确实需要指定轴。使用axis=1，我的lambda函数中的x代表数据row的每一行。|1681176|最后，我还忘记提到，通过执行以下操作，您可以更快地使用虚拟变量完成这一任务：|1681177|per_country['virus_confirmed']+=+per_country['cases']*per_country['confirmed']|1681178|因为未确认的列被标记为0，所以它将给出相同的结果，但以向量化的方式。如果您正在使用新冠肺炎数据集，我认为您不会注意到有什么很大的不同，但寻找机会进行矢量化是一个好习惯。|1681179|entityMap^0|4|5|D|J|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|11|8|@]|D|@]|E|$]]|$1|O|3|P|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|Q|3|R|5|6|7|13|8|@]|D|@]|E|$]]|$1|S|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|T|$]]

When you use <code>apply</code>, it's implicitly passing each record of <code>per_country['type']</code> to func. If you want to do this more simply and clearer, you can use a lambda function.

<pre><code>per_country['virus_confirmed'] = per_country.apply(lambda x: x['cases'] if x['type']=="confirmed" else 0, axis=1)
</code></pre>

EDIT:
For the apply function, note that you're applying it to a series in the DataFrame. This means that x is each record from that column and that you don't need to specify the axis. I applied it to the whole DataFrame, which means I do need to specify the axis. With axis=1, x in my lambda function stands in for each row of the dataframe.

Lastly, I also forgot to mention that you can get this done even quicker with the dummy variables by doing:

<pre><code>per_country['virus_confirmed'] = per_country['cases']*per_country['confirmed']
</code></pre>

Because columns that aren't confirmed are marked with a 0, it will give you the same result, but in a vectorized manner. If you're working with a Covid-19 dataset, I don't think you'll notice a huge difference, but it's a good habit to look for opportunities to vectorize.

blocks|key|669991|text|func是一个接受0参数的函数。apply将数据传递给函数的第一个参数。这显然失败了，因为func不接受任何参数。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|669992|您似乎对Python范围的基本知识感到困惑。您可以向here咨询介绍。|669993|简而言之，您正在尝试编写一个应用于数据帧的每一行的函数。因此，函数func必须接受一个行作为它的参数。|669994|#+Each+row+will+be+passed+as+the+"x"+parameter

def+func(x):
++++if+x['confirmed']+==+'confirmed':
++++++++return+x['cases']
++++else:
++++++++return+0

#+axis=1+is+required+to+apply+the+function+row-wise

per_country['virus_confirmed']+=+per_country.apply(func,+axis=1)|code-block|syntax|javascript|669995|也就是说，你要问的函数似乎与你显示的数据无关，所以这个问题有点让人困惑。|669996|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/291978/short-description-of-the-scoping-rules#292502^0|0|4|G|5|19|4|0|Q|4|0|0|X|4|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|15|8|@]|D|@$9|16|A|17|1|18]]|E|$]]|$1|H|3|I|5|6|7|19|8|@$9|1A|A|1B|B|C]]|D|@]|E|$]]|$1|J|3|K|5|L|7|1C|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|1D|8|@]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]]]

<code>func</code> is a function that takes 0 parameters. <code>apply</code> passes data to the first parameter of the function. This obviously fails because <code>func</code> does not accept any parameters.

You seem confused about the basics of Python scoping. You can consult <a href="https://stackoverflow.com/questions/291978/short-description-of-the-scoping-rules#292502">here</a> for an introduction.

In short, you are trying to write a function that is applied to each row of the data frame. Therefore your function <code>func</code> must accept a row as its parameter.

<pre class="lang-py prettyprint-override"><code># Each row will be passed as the "x" parameter

def func(x):
 if x['confirmed'] == 'confirmed':
 return x['cases']
 else:
 return 0

# axis=1 is required to apply the function row-wise

per_country['virus_confirmed'] = per_country.apply(func, axis=1)
</code></pre>

That said, the function you're asking about doesn't seem related to the data you're showing, so this whole question is a bit confusing.

blocks|key|112924|text|另一个注意事项:在类别之间不存在任何关系的情况下，一个热编码适合于分类数据。它包括用二进制向量表示每个分类变量，该向量对每个唯一的标签都有一个元素，并使用1和所有其他元素(也就是您正在尝试的元素)标记类标签。scikit-learn库提供了自动对一个或多个变量进行热编码的OneHotEncoder。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|112925|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Just on another note : A one hot encoding is appropriate for categorical data where no relationship exists between categories. It involves representing each categorical variable with a binary vector that has one element for each unique label and marking the class label with a 1 and all other elements 0 which is the one you are trying. The scikit-learn library provides the OneHotEncoder to automatically one hot encode one or more variables.

I have the following data frame:

<img src="https://i.stack.imgur.com/TRrbe.png" alt="Data frame screenshot">

and I want to convert it to this format:

<img src="https://i.stack.imgur.com/RTRMW.png" alt="enter image description here">

I have succeed that with <code>get_dummies</code> but i am trying to do the same with defining a function like:

<pre><code>def func():
 if per_country['confirmed']=='confirmed':
 return per_country['cases']
 else:
 return 0

per_country['virus_confirmed']=per_country['type'].apply(func)
</code></pre>

The <code>per_country</code> dataframe is the first screenshot.

But i am getting this error:

<img src="https://i.stack.imgur.com/HCdwx.png" alt="enter image description here">

What am i doing wrong?

Split categorical column to multiple columns in Python?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有以下数据框架：​​我想把它转换成这种格式：​​我已经在get_dummies中取得了成功，但我也试图在定义如下函数时做同样的工作：def func():    if per_country['confirmed']=='confirmed':        return per_country['cases']  ...

问在Python中将分类列拆分为多列？
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