使用多个 If-else 创建 Pandas 变量
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Pandas variable creation using multiple If-else
提问by Baktaawar
Need help with Pandas multiple IF-ELSE statements. I have a test dataset (titanic) as follows:
需要 Pandas 多个 IF-ELSE 语句的帮助。我有一个测试数据集（泰坦尼克号）如下：
ID Survived Pclass Name Sex Age
1 0 3 Braund male 22
2 1 1 Cumings, Mrs. female 38
3 1 3 Heikkinen, Miss. Laina female 26
4 1 1 Futrelle, Mrs. female 35
5 0 3 Allen, Mr. male 35
6 0 3 Moran, Mr. male
7 0 1 McCarthy, Mr. male 54
8 0 3 Palsson, Master male 2
where Id is the passenger id. I want to create a new flag variable in this data frame which has the following rule:
其中 Id 是乘客 ID。我想在这个数据框中创建一个新的标志变量，它具有以下规则：
if Sex=="female" or (Pclass==1 and Age <18) then 1 else 0.
Now to do this I tried a few approaches. This is how I approached first:
现在要做到这一点，我尝试了一些方法。这是我第一次接触的方式：
df=pd.read_csv(data.csv)
for passenger_index,passenger in df.iterrows():
if passenger['Sex']=="female" or (passenger['Pclass']==1 and passenger['Age']<18):
df['Prediction']=1
else:
df['Prediction']=0
The problem with above code is that it creates a Prediction variable in df but with all values as 0.
上面代码的问题在于它在 df 中创建了一个预测变量，但所有值都为 0。
However if I use the same code but instead output it to a dictionary it gives the right answer as shown below:
但是，如果我使用相同的代码而是将其输出到字典，它会给出正确的答案，如下所示：
prediction={}
df=pd.read_csv(data.csv)
for passenger_index,passenger in df.iterrows():
if passenger['Sex']=="female" or (passenger['Pclass']==1 and passenger['Age']<18):
prediction[passenger['ID']=1
else:
prediction[passenger['ID']=0
This gives a dict prediction with keys as ID and values as 1 or 0 based on the above logic.
根据上述逻辑，这给出了一个以键为 ID、值为 1 或 0 的字典预测。
So why the df variable works wrongly?. I even tried by first defining a function and then calling it. Gave the same ans as first.
那么为什么 df 变量工作错误呢？我什至尝试先定义一个函数，然后再调用它。给出与第一相同的答案。
So, how can we do this in pandas?.
那么，我们如何在Pandas中做到这一点？
Secondly, I guess the same can be done if we can just use some multiple if-else statements. I know np.where but it is not allowing to add 'and' condition. So here is what I was trying:
其次，我想如果我们可以使用多个 if-else 语句，也可以做到同样的事情。我知道 np.where 但它不允许添加“和”条件。所以这就是我正在尝试的：
df['Prediction']=np.where(df['Sex']=="female",1,np.where((df['Pclass']==1 and df['Age']<18),1,0)
The above gave an error for 'and' keyword in where.
上面在 where 中给出了“and”关键字的错误。
So can someone help?. Solutions with multiple approache using np.where(simple if-else like) and using some function(applymap etc) or modifications to what I wrote earlier would be really appreciated.
所以有人可以帮忙吗？使用 np.where（简单的 if-else 之类）并使用一些函数（applymap 等）或对我之前编写的内容进行修改的多种方法的解决方案将不胜感激。
Also how do we do the same using some applymap or apply/map method of df?.
另外我们如何使用df的一些applymap或apply/map方法来做同样的事情？。
回答by unutbu
Instead of looping through the rows using df.iterrows(which is relatively slow), you can assign the desired values to the Predictioncolumn in one assignment:
df.iterrows您可以Prediction在一次分配中将所需的值分配给列，而不是使用循环遍历行（这相对较慢）：
In [27]: df['Prediction'] = ((df['Sex']=='female') | ((df['Pclass']==1) & (df['Age']<18))).astype('int')
In [29]: df['Prediction']
Out[29]:
0 0
1 1
2 1
3 1
4 0
5 0
6 0
7 0
Name: Prediction, dtype: int32
For your first approach, remember that df['Prediction']represents an entire column of df, so df['Prediction']=1assigns the value 1 to each row in that column. Since df['Prediction']=0was the last assignment, the entire column ended up being filled with zeros.
对于您的第一种方法，请记住df['Prediction']代表 的一整列df，因此df['Prediction']=1将值 1 分配给该列中的每一行。由于df['Prediction']=0是最后一次分配，整个列最终都被零填充。
For your second approach, note that you need to use ¬ andto perform an elementwiselogical-and operation on two NumPy arrays or Pandas NDFrames. Thus, you could use
对于您的第二种方法，请注意您需要使用¬对两个 NumPy 数组或 Pandas NDFrameand执行元素逻辑与运算。因此，您可以使用
In [32]: np.where(df['Sex']=='female', 1, np.where((df['Pclass']==1)&(df['Age']<18), 1, 0))
Out[32]: array([0, 1, 1, 1, 0, 0, 0, 0])
though I think it is much simpler to just use |for logical-or and &for logical-and:
尽管我认为仅|用于逻辑或和&逻辑与要简单得多：
In [34]: ((df['Sex']=='female') | ((df['Pclass']==1) & (df['Age']<18)))
Out[34]:
0 False
1 True
2 True
3 True
4 False
5 False
6 False
7 False
dtype: bool
