Using Apply in Pandas Lambda functions with multiple if statements

Question

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.

Answer 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

Answer 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)

Answer 3

The apply lambda function actually does the job here, I just wonder what the problem was.... as your syntax looks ok and it works....

df1= [80000, 8000000, 8000000000, 800000000000]
df=pd.DataFrame(df1)
df.columns=['size']
df['Classification']=df['size'].apply(lambda x: '<1m' if x<1000000  else '1-10m' if 1000000<x<10000000 else '1bi')
df

Output:

Answer 4

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.