In pandas, is inplace = True considered harmful, or not?

In pandas, is inplace = True considered harmful, or not?

Yes, it is. Not just harmful. Quite harmful. This GitHub issue is proposing the inplace argument be deprecated api-wide sometime in the near future. In a nutshell, here’s everything wrong with the inplace argument:

• inplace, contrary to what the name implies, often does not prevent copies from being created, and (almost) never offers any performance benefits
• inplace does not work with method chaining
• inplace can lead to the dreaded SettingWithCopyWarning when called on a DataFrame column, and may sometimes fail to update the column in-place

The pain points above are all common pitfall for beginners, so removing this option will simplify the API greatly.

We take a look at the points above in more depth.

Performance
It is a common misconception that using inplace=True will lead to more efficient or optimized code. In general, there are no performance benefits to using inplace=True (but there are rare exceptions which are mostly a result of implementation detail in the library and should not be used as a crutch to advocate for this argument’s usage). Most in-place and out-of-place versions of a method create a copy of the data anyway, with the in-place version automatically assigning the copy back. The copy cannot be avoided.

Method Chaining
inplace=True also hinders method chaining. Contrast the working of

result = df.some_function1().reset_index().some_function2()

As opposed to

temp = df.some_function1()
temp.reset_index(inplace=True)
result = temp.some_function2()

Unintended Pitfalls
One final caveat to keep in mind is that calling inplace=True can trigger the SettingWithCopyWarning:

df = pd.DataFrame({'a': [3, 2, 1], 'b': ['x', 'y', 'z']})

df2 = df[df['a'] > 1]
df2['b'].replace({'x': 'abc'}, inplace=True)
# SettingWithCopyWarning: 
# A value is trying to be set on a copy of a slice from a DataFrame

Which can cause unexpected behavior.