why should I make a copy of a data frame in pandas

Solution 1:

This expands on Paul's answer. In Pandas, indexing a DataFrame returns a reference to the initial DataFrame. Thus, changing the subset will change the initial DataFrame. Thus, you'd want to use the copy if you want to make sure the initial DataFrame shouldn't change. Consider the following code:

df = DataFrame({'x': [1,2]})
df_sub = df[0:1]
df_sub.x = -1
print(df)

You'll get:

x
0 -1
1  2

In contrast, the following leaves df unchanged:

df_sub_copy = df[0:1].copy()
df_sub_copy.x = -1

Solution 2:

Because if you don't make a copy then the indices can still be manipulated elsewhere even if you assign the dataFrame to a different name.

For example:

df2 = df
func1(df2)
func2(df)

func1 can modify df by modifying df2, so to avoid that:

df2 = df.copy()
func1(df2)
func2(df)

Solution 3:

It's necessary to mention that returning copy or view depends on kind of indexing.

The pandas documentation says:

Returning a view versus a copy

The rules about when a view on the data is returned are entirely dependent on NumPy. Whenever an array of labels or a boolean vector are involved in the indexing operation, the result will be a copy. With single label / scalar indexing and slicing, e.g. df.ix[3:6] or df.ix[:, 'A'], a view will be returned.

Solution 4:

The primary purpose is to avoid chained indexing and eliminate the SettingWithCopyWarning.

Here chained indexing is something like dfc['A'][0] = 111

The document said chained indexing should be avoided in Returning a view versus a copy. Here is a slightly modified example from that document:

In [1]: import pandas as pd

In [2]: dfc = pd.DataFrame({'A':['aaa','bbb','ccc'],'B':[1,2,3]})

In [3]: dfc
Out[3]:
    A   B
0   aaa 1
1   bbb 2
2   ccc 3

In [4]: aColumn = dfc['A']

In [5]: aColumn[0] = 111
SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame

In [6]: dfc
Out[6]:
    A   B
0   111 1
1   bbb 2
2   ccc 3

Here the aColumn is a view and not a copy from the original DataFrame, so modifying aColumn will cause the original dfc be modified too. Next, if we index the row first:

In [7]: zero_row = dfc.loc[0]

In [8]: zero_row['A'] = 222
SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame

In [9]: dfc
Out[9]:
    A   B
0   111 1
1   bbb 2
2   ccc 3

This time zero_row is a copy, so the original dfc is not modified.

From these two examples above, we see it's ambiguous whether or not you want to change the original DataFrame. This is especially dangerous if you write something like the following:

In [10]: dfc.loc[0]['A'] = 333
SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame

In [11]: dfc
Out[11]:
    A   B
0   111 1
1   bbb 2
2   ccc 3

This time it didn't work at all. Here we wanted to change dfc, but we actually modified an intermediate value dfc.loc[0] that is a copy and is discarded immediately. It’s very hard to predict whether the intermediate value like dfc.loc[0] or dfc['A'] is a view or a copy, so it's not guaranteed whether or not original DataFrame will be updated. That's why chained indexing should be avoided, and pandas generates the SettingWithCopyWarning for this kind of chained indexing update.

Now is the use of .copy(). To eliminate the warning, make a copy to express your intention explicitly:

In [12]: zero_row_copy = dfc.loc[0].copy()

In [13]: zero_row_copy['A'] = 444 # This time no warning

Since you are modifying a copy, you know the original dfc will never change and you are not expecting it to change. Your expectation matches the behavior, then the SettingWithCopyWarning disappears.

Note, If you do want to modify the original DataFrame, the document suggests you use loc:

In [14]: dfc.loc[0,'A'] = 555

In [15]: dfc
Out[15]:
    A   B
0   555 1
1   bbb 2
2   ccc 3

Solution 5:

Assumed you have data frame as below

df1
     A    B    C    D
4 -1.0 -1.0 -1.0 -1.0
5 -1.0 -1.0 -1.0 -1.0
6 -1.0 -1.0 -1.0 -1.0
6 -1.0 -1.0 -1.0 -1.0

When you would like create another df2 which is identical to df1, without copy

df2=df1
df2
     A    B    C    D
4 -1.0 -1.0 -1.0 -1.0
5 -1.0 -1.0 -1.0 -1.0
6 -1.0 -1.0 -1.0 -1.0
6 -1.0 -1.0 -1.0 -1.0

And would like modify the df2 value only as below

df2.iloc[0,0]='changed'

df2
         A    B    C    D
4  changed -1.0 -1.0 -1.0
5       -1 -1.0 -1.0 -1.0
6       -1 -1.0 -1.0 -1.0
6       -1 -1.0 -1.0 -1.0

At the same time the df1 is changed as well

df1
         A    B    C    D
4  changed -1.0 -1.0 -1.0
5       -1 -1.0 -1.0 -1.0
6       -1 -1.0 -1.0 -1.0
6       -1 -1.0 -1.0 -1.0

Since two df as same object, we can check it by using the id

id(df1)
140367679979600
id(df2)
140367679979600

So they as same object and one change another one will pass the same value as well.

If we add the copy, and now df1 and df2 are considered as different object, if we do the same change to one of them the other will not change.

df2=df1.copy()
id(df1)
140367679979600
id(df2)
140367674641232

df1.iloc[0,0]='changedback'
df2
         A    B    C    D
4  changed -1.0 -1.0 -1.0
5       -1 -1.0 -1.0 -1.0
6       -1 -1.0 -1.0 -1.0
6       -1 -1.0 -1.0 -1.0

Good to mention, when you subset the original dataframe, it is safe to add the copy as well in order to avoid the SettingWithCopyWarning