How are iloc and loc different?

Can someone explain how these two methods of slicing are different? I've seen the docs
and I've seen previous similar questions (1, 2), but I still find myself unable to understand how they are different. To me, they seem interchangeable in large part, because they are at the lower levels of slicing.

For example, say we want to get the first five rows of a DataFrame. How is it that these two work?
df.loc[:5]
df.iloc[:5]


Can someone present cases where the distinction in uses are clearer?

Once upon a time, I also wanted to know how these two functions differed from df.ix[:5] but ix has been removed from pandas 1.0, so I don't care anymore.

Label vs. Location

The main distinction between the two methods is:

-
loc gets rows (and/or columns) with particular labels.

-
iloc gets rows (and/or columns) at integer locations.

To demonstrate, consider a series s of characters with a non-monotonic integer index:

>>> s = pd.Series(list("abcdef"), index=[49, 48, 47, 0, 1, 2]) 
49    a
48    b
47    c
0     d
1     e
2     f

>>> s.loc[0]    # value at index label 0
'd'

>>> s.iloc[0]   # value at index location 0
'a'

>>> s.loc[0:1]  # rows at index labels between 0 and 1 (inclusive)
0    d
1    e

>>> s.iloc[0:1] # rows at index location between 0 and 1 (exclusive)
49    a

Here are some of the differences/similarities between s.loc and s.iloc when passed various objects:

description
s.loc[]
s.iloc[]

0
single item
Value at index label 0 (the string 'd')
Value at index location 0 (the string 'a')

0:1
slice
Two rows (labels 0 and 1)
One row (first row at location 0)

1:47
slice with out-of-bounds end
Zero rows (empty Series)
Five rows (location 1 onwards)

1:47:-1
slice with negative step
three rows (labels 1 back to 47)
Zero rows (empty Series)

[2, 0]
integer list
Two rows with given labels
Two rows with given locations

s > 'e'
Bool series (indicating which values have the property)
One row (containing 'f')
NotImplementedError

(s>'e').values
Bool array
One row (containing 'f')
Same as loc

999
int object not in index
KeyError
IndexError (out of bounds)

-1
int object not in index
KeyError
Returns last value in s

lambda x: x.index[3]
callable applied to series (here returning 3rd item in index)
s.loc[s.index[3]]
s.iloc[s.index[3]]

loc's label-querying capabilities extend well-beyond integer indexes and it's worth highlighting a couple of additional examples.

Here's a Series where the index contains string objects:

>>> s2 = pd.Series(s.index, index=s.values)
>>> s2
a    49
b    48
c    47
d     0
e     1
f     2

Since loc is label-based, it can fetch the first value in the Series using s2.loc['a']. It can also slice with non-integer objects:

>>> s2.loc['c':'e']  # all rows lying between 'c' and 'e' (inclusive)
c    47
d     0
e     1

For DateTime indexes, we don't need to pass the exact date/time to fetch by label. For example:

>>> s3 = pd.Series(list('abcde'), pd.date_range('now', periods=5, freq='M')) 
>>> s3
2021-01-31 16:41:31.879768    a
2021-02-28 16:41:31.879768    b
2021-03-31 16:41:31.879768    c
2021-04-30 16:41:31.879768    d
2021-05-31 16:41:31.879768    e

Then to fetch the row(s) for March/April 2021 we only need:

>>> s3.loc['2021-03':'2021-04']
2021-03-31 17:04:30.742316    c
2021-04-30 17:04:30.742316    d

Rows and Columns

loc and iloc work the same way with DataFrames as they do with Series. It's useful to note that both methods can address columns and rows together.

When given a tuple, the first element is used to index the rows and, if it exists, the second element is used to index the columns.

Consider the DataFrame defined below:

>>> import numpy as np 
>>> df = pd.DataFrame(np.arange(25).reshape(5, 5),  
                      index=list('abcde'), 
                      columns=['x','y','z', 8, 9])
>>> df
    x   y   z   8   9
a   0   1   2   3   4
b   5   6   7   8   9
c  10  11  12  13  14
d  15  16  17  18  19
e  20  21  22  23  24

Then for example:

>>> df.loc['c': , :'z']  # rows 'c' and onwards AND columns up to 'z'
    x   y   z
c  10  11  12
d  15  16  17
e  20  21  22

>>> df.iloc[:, 3]        # all rows, but only the column at index location 3
a     3
b     8
c    13
d    18
e    23

Sometimes we want to mix label and positional indexing methods for the rows and columns, somehow combining the capabilities of loc and iloc.

For example, consider the following DataFrame. How best to slice the rows up to and including 'c' and take the first four columns?

>>> import numpy as np 
>>> df = pd.DataFrame(np.arange(25).reshape(5, 5),  
                      index=list('abcde'), 
                      columns=['x','y','z', 8, 9])
>>> df
    x   y   z   8   9
a   0   1   2   3   4
b   5   6   7   8   9
c  10  11  12  13  14
d  15  16  17  18  19
e  20  21  22  23  24

We can achieve this result using iloc and the help of another method:

>>> df.iloc[:df.index.get_loc('c') + 1, :4]
    x   y   z   8
a   0   1   2   3
b   5   6   7   8
c  10  11  12  13

get_loc() is an index method meaning "get the position of the label in this index". Note that since slicing with iloc is exclusive of its endpoint, we must add 1 to this value if we want row 'c' as well.

Stack Overflow Q#31593201, score: 1608