问题
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
...
I want it to look like this:
Size Classification
1 80000 <1m
2 8000000 1-10m
3 8000000000 >1bi
...
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 ...)
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.
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)
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
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.
回答1:
Here is a small example that you can build upon:
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)
Returns:
Size Classification
0 80000 <1m
1 8000000 1-10m
2 800000000 N/A
回答2:
You can use pd.cut function:
bins = [0, 1000000, 10000000, 50000000, ...]
labels = ['<1m','1-10m','10-50m', ...]
df['Classification'] = pd.cut(df['Size'], bins=bins, labels=labels)
回答3:
Using Numpy's searchsorted
labels = np.array(['<1m', '1-10m', '10-50m', '>50m'])
bins = np.array([1E6, 1E7, 5E7])
# Using assign is my preference as it produces a copy of df with new column
df.assign(Classification=labels[bins.searchsorted(df['Size'].values)])
If you wanted to produce new column in existing dataframe
df['Classification'] = labels[bins.searchsorted(df['Size'].values)]
Some Explanation
From Docs:np.searchsorted
Find indices where elements should be inserted to maintain order.
Find the indices into a sorted array a such that, if the corresponding elements in v were inserted before the indices, the order of a would be preserved.
The labels array has a length greater than that of bins by one. Because when something is greater than the maximum value in bins, searchsorted returns a -1. When we slice labels this grabs the last label.
来源:https://stackoverflow.com/questions/48872234/using-apply-in-pandas-lambda-functions-with-multiple-if-statements