在带有多个 if 语句的 Pandas Lambda 函数中使用 Apply [英] Using Apply in Pandas Lambda functions with multiple if statements

问题描述

我正在尝试根据这样的数据框中人的大小来推断分类:

I'm trying to infer a classification according to the size of a person in a dataframe like this one:

      Size
1     80000
2     8000000
3     8000000000
...

我希望它看起来像这样:

I want it to look like this:

      Size        Classification
1     80000       <1m
2     8000000     1-10m
3     8000000000  >1bi
...

我知道理想的过程是应用这样的 lambda 函数:

I understand that the ideal process would be to apply a lambda function like this:

df['Classification']=df['Size'].apply(lambda x: "<1m" if x<1000000 else "1-10m" if 1000000<x<10000000 else ...)

我查看了一些关于 lambda 函数中多个 if 的帖子，这里是一个示例链接，但由于某种原因，该 synthax 在多个 ifs 语句中对我不起作用，但它在单个 if 条件下工作.

I checked a few posts regarding multiple ifs in a lambda function, here is an example link, but that synthax is not working for me for some reason in a multiple ifs statement, but it was working in a single if condition.

所以我尝试了这个非常优雅"的解决方案:

So I tried this "very elegant" solution:

df['Classification']=df['Size'].apply(lambda x: "<1m" if x<1000000 else pass)
df['Classification']=df['Size'].apply(lambda x: "1-10m" if 1000000 < x < 10000000 else pass)
df['Classification']=df['Size'].apply(lambda x: "10-50m" if 10000000 < x < 50000000 else pass)
df['Classification']=df['Size'].apply(lambda x: "50-100m" if 50000000 < x < 100000000 else pass)
df['Classification']=df['Size'].apply(lambda x: "100-500m" if 100000000 < x < 500000000 else pass)
df['Classification']=df['Size'].apply(lambda x: "500m-1bi" if 500000000 < x < 1000000000 else pass)
df['Classification']=df['Size'].apply(lambda x: ">1bi" if 1000000000 < x else pass)

发现pass"似乎也不适用于 lambda 函数:

Works out that "pass" seems not to apply to lambda functions as well:

df['Classification']=df['Size'].apply(lambda x: "<1m" if x<1000000 else pass)
SyntaxError: invalid syntax

对于 Pandas 的 apply 方法中的 lambda 函数内的多个 if 语句的正确合成有任何建议吗?多线或单线解决方案都适合我.

Any suggestions on the correct synthax for a multiple if statement inside a lambda function in an apply method in Pandas? Either multi-line or single line solutions work for me.

推荐答案

以下是一个小示例，您可以在此基础上进行构建:

Here is a small example that you can build upon:

基本上，lambda x: x.. 是一个函数的短代码.apply 真正需要的是一个您可以轻松重新创建自己的功能.

Basically, lambda x: x.. is the short one-liner of a function. What apply really asks for is a function which you can easily recreate yourself.

import pandas as pd

# Recreate the dataframe
data = dict(Size=[80000,8000000,800000000])
df = pd.DataFrame(data)

# Create a function that returns desired values
# You only need to check upper bound as the next elif-statement will catch the value
def func(x):
    if x < 1e6:
        return "<1m"
    elif x < 1e7:
        return "1-10m"
    elif x < 5e7:
        return "10-50m"
    else:
        return 'N/A'
    # Add elif statements....

df['Classification'] = df['Size'].apply(func)

print(df)

返回:

        Size Classification
0      80000            <1m
1    8000000          1-10m
2  800000000            N/A

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