Fix Python – python pandas dataframe, is it pass-by-value or pass-by-reference


Asked By – nos

If I pass a dataframe to a function and modify it inside the function, is it pass-by-value or pass-by-reference?

I run the following code

a = pd.DataFrame({'a':[1,2], 'b':[3,4]})
def letgo(df):
    df = df.drop('b',axis=1)

the value of a does not change after the function call. Does it mean it is pass-by-value?

I also tried the following

xx = np.array([[1,2], [3,4]])
def letgo2(x):
    x[1,1] = 100
def letgo3(x):
    x = np.array([[3,3],[3,3]])

It turns out letgo2() does change xx and letgo3() does not. Why is it like this?

Now we will see solution for issue: python pandas dataframe, is it pass-by-value or pass-by-reference


The short answer is, Python always does pass-by-value, but every Python variable is actually a pointer to some object, so sometimes it looks like pass-by-reference.

In Python every object is either mutable or non-mutable. e.g., lists, dicts, modules and Pandas data frames are mutable, and ints, strings and tuples are non-mutable. Mutable objects can be changed internally (e.g., add an element to a list), but non-mutable objects cannot.

As I said at the start, you can think of every Python variable as a pointer to an object. When you pass a variable to a function, the variable (pointer) within the function is always a copy of the variable (pointer) that was passed in. So if you assign something new to the internal variable, all you are doing is changing the local variable to point to a different object. This doesn’t alter (mutate) the original object that the variable pointed to, nor does it make the external variable point to the new object. At this point, the external variable still points to the original object, but the internal variable points to a new object.

If you want to alter the original object (only possible with mutable data types), you have to do something that alters the object without assigning a completely new value to the local variable. This is why letgo() and letgo3() leave the external item unaltered, but letgo2() alters it.

As @ursan pointed out, if letgo() used something like this instead, then it would alter (mutate) the original object that df points to, which would change the value seen via the global a variable:

def letgo(df):
    df.drop('b', axis=1, inplace=True)

a = pd.DataFrame({'a':[1,2], 'b':[3,4]})
letgo(a)  # will alter a

In some cases, you can completely hollow out the original variable and refill it with new data, without actually doing a direct assignment, e.g. this will alter the original object that v points to, which will change the data seen when you use v later:

def letgo3(x):
    x[:] = np.array([[3,3],[3,3]])

v = np.empty((2, 2))
letgo3(v)   # will alter v

Notice that I’m not assigning something directly to x; I’m assigning something to the entire internal range of x.

If you absolutely must create a completely new object and make it visible externally (which is sometimes the case with pandas), you have two options. The ‘clean’ option would be just to return the new object, e.g.,

def letgo(df):
    df = df.drop('b',axis=1)
    return df

a = pd.DataFrame({'a':[1,2], 'b':[3,4]})
a = letgo(a)

Another option would be to reach outside your function and directly alter a global variable. This changes a to point to a new object, and any function that refers to a afterward will see that new object:

def letgo():
    global a
    a = a.drop('b',axis=1)

a = pd.DataFrame({'a':[1,2], 'b':[3,4]})
letgo()   # will alter a!

Directly altering global variables is usually a bad idea, because anyone who reads your code will have a hard time figuring out how a got changed. (I generally use global variables for shared parameters used by many functions in a script, but I don’t let them alter those global variables.)

This question is answered By – Matthias Fripp

This answer is collected from stackoverflow and reviewed by FixPython community admins, is licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0