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Active reading - including complying with the Jon Skeet Decree - <https://twitter.com/PeterMortensen/status/976400000942034944>.
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Peter Mortensen
Peter Mortensen

l[start:end]

l[start:end]

where l is some collection, start is an inclusive index, and end is an exclusive index.

In [1]: l = list(range(10))

In [2]: l[:5] # firstFirst five elements
Out[2]: [0, 1, 2, 3, 4]

In [3]: l[-5:] # lastLast five elements
Out[3]: [5, 6, 7, 8, 9]

When slicing from the start, you can omit the zero index, and when slicing to the end, you can omit the final index since it is redundant, so do not be verbose:

In [7]: l[:-1] # includeInclude all elements but the last one
Out[7]: [0, 1, 2, 3, 4, 5, 6, 7, 8]

In [8]: l[-3:] # takeTake the last 3three elements
Out[8]: [7, 8, 9]

In [9]: l[:20] # 20 is out of index bounds, and l[20] will raise an IndexError exception
Out[9]: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

In [11]: l[-20:] # -20 is out of index bounds, and l[-20] will raise an IndexError exception
Out[11]: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

Keep in mind that the result of slicing a collection is a whole new collection. In addition, when using slice notation in assignments, the length of the slice assignmentassignments do not need to be the same. The values before and after the assigned slice will be kept, and the collection will shrink or grow to contain the new values:

In [16]: l[2:6] = list('abc') # assigningAssigning lessfewer elements than the ones contained in the sliced collection l[2:6]

In [17]: l
Out[17]: [0, 1, 'a', 'b', 'c', 6, 7, 8, 9]

In [18]: l[2:5] = list('hello') # assigningAssigning more elements than the ones contained in the sliced collection l [2:5]

In [19]: l
Out[19]: [0, 1, 'h', 'e', 'l', 'l', 'o', 6, 7, 8, 9]

In [22]: l = list(range(10))

In [23]: l[::2] # takeTake the elements which indexes are even
Out[23]: [0, 2, 4, 6, 8]

In [24]: l[1::2] # takeTake the elements which indexes are odd
Out[24]: [1, 3, 5, 7, 9]

However, using a negative value for step could become very confusing. Moreover, in order to be PythonicPythonic, you should avoid using start, end, and step in a single slice. In case this is required, consider doing this in two assignments (one to slice, and the other to stride).

In [29]: l = l[::2] # thisThis step is for striding

In [30]: l
Out[30]: [0, 2, 4, 6, 8]

In [31]: l = l[1:-1] # thisThis step is for slicing

In [32]: l
Out[32]: [2, 4, 6]

l[start:end]

where l is some collection, start is an inclusive index and end is an exclusive index.

In [1]: l = list(range(10))

In [2]: l[:5] # first five elements
Out[2]: [0, 1, 2, 3, 4]

In [3]: l[-5:] # last five elements
Out[3]: [5, 6, 7, 8, 9]

When slicing from start, you can omit the zero index, and when slicing to the end, you can omit the final index since it is redundant, so do not be verbose:

In [7]: l[:-1] # include all elements but the last one
Out[7]: [0, 1, 2, 3, 4, 5, 6, 7, 8]

In [8]: l[-3:] # take the last 3 elements
Out[8]: [7, 8, 9]

In [9]: l[:20] # 20 is out of index bounds, l[20] will raise an IndexError exception
Out[9]: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

In [11]: l[-20:] # -20 is out of index bounds, l[-20] will raise an IndexError exception
Out[11]: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

Keep in mind that the result of slicing a collection is a whole new collection. In addition, when using slice notation in assignments, the length of the slice assignment do not need to be the same. The values before and after the assigned slice will be kept, and the collection will shrink or grow to contain the new values:

In [16]: l[2:6] = list('abc') # assigning less elements than the ones contained in the sliced collection l[2:6]

In [17]: l
Out[17]: [0, 1, 'a', 'b', 'c', 6, 7, 8, 9]

In [18]: l[2:5] = list('hello') # assigning more elements than the ones contained in the sliced collection l [2:5]

In [19]: l
Out[19]: [0, 1, 'h', 'e', 'l', 'l', 'o', 6, 7, 8, 9]

In [22]: l = list(range(10))

In [23]: l[::2] # take the elements which indexes are even
Out[23]: [0, 2, 4, 6, 8]

In [24]: l[1::2] # take the elements which indexes are odd
Out[24]: [1, 3, 5, 7, 9]

However, using a negative value for step could become very confusing. Moreover, in order to be Pythonic, you should avoid using start, end, and step in a single slice. In case this is required, consider doing this in two assignments (one to slice, and the other to stride).

In [29]: l = l[::2] # this step is for striding

In [30]: l
Out[30]: [0, 2, 4, 6, 8]

In [31]: l = l[1:-1] # this step is for slicing

In [32]: l
Out[32]: [2, 4, 6]

l[start:end]

where l is some collection, start is an inclusive index, and end is an exclusive index.

In [1]: l = list(range(10))

In [2]: l[:5] # First five elements
Out[2]: [0, 1, 2, 3, 4]

In [3]: l[-5:] # Last five elements
Out[3]: [5, 6, 7, 8, 9]

When slicing from the start, you can omit the zero index, and when slicing to the end, you can omit the final index since it is redundant, so do not be verbose:

In [7]: l[:-1] # Include all elements but the last one
Out[7]: [0, 1, 2, 3, 4, 5, 6, 7, 8]

In [8]: l[-3:] # Take the last three elements
Out[8]: [7, 8, 9]

In [9]: l[:20] # 20 is out of index bounds, and l[20] will raise an IndexError exception
Out[9]: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

In [11]: l[-20:] # -20 is out of index bounds, and l[-20] will raise an IndexError exception
Out[11]: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

Keep in mind that the result of slicing a collection is a whole new collection. In addition, when using slice notation in assignments, the length of the slice assignments do not need to be the same. The values before and after the assigned slice will be kept, and the collection will shrink or grow to contain the new values:

In [16]: l[2:6] = list('abc') # Assigning fewer elements than the ones contained in the sliced collection l[2:6]

In [17]: l
Out[17]: [0, 1, 'a', 'b', 'c', 6, 7, 8, 9]

In [18]: l[2:5] = list('hello') # Assigning more elements than the ones contained in the sliced collection l [2:5]

In [19]: l
Out[19]: [0, 1, 'h', 'e', 'l', 'l', 'o', 6, 7, 8, 9]

In [22]: l = list(range(10))

In [23]: l[::2] # Take the elements which indexes are even
Out[23]: [0, 2, 4, 6, 8]

In [24]: l[1::2] # Take the elements which indexes are odd
Out[24]: [1, 3, 5, 7, 9]

However, using a negative value for step could become very confusing. Moreover, in order to be Pythonic, you should avoid using start, end, and step in a single slice. In case this is required, consider doing this in two assignments (one to slice, and the other to stride).

In [29]: l = l[::2] # This step is for striding

In [30]: l
Out[30]: [0, 2, 4, 6, 8]

In [31]: l = l[1:-1] # This step is for slicing

In [32]: l
Out[32]: [2, 4, 6]
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lmiguelvargasf
lmiguelvargasf

In Python, the most basic form for slicing is the following:

l[start:end]

where l is some collection, start is an inclusive index and end is an exclusive index.

In [1]: l = list(range(10))

In [2]: l[:5] # first five elements
Out[2]: [0, 1, 2, 3, 4]

In [3]: l[-5:] # last five elements
Out[3]: [5, 6, 7, 8, 9]

When slicing from start, you can omit the zero index, and when slicing to the end, you can omit the final index since it is redundant, so do not be verbose:

In [5]: l[:3] == l[0:3]
Out[5]: True

In [6]: l[7:] == l[7:len(l)]
Out[6]: True

Negative integers are useful when doing offsets relative to the end of a collection:

In [7]: l[:-1] # include all elements but the last one
Out[7]: [0, 1, 2, 3, 4, 5, 6, 7, 8]

In [8]: l[-3:] # take the last 3 elements
Out[8]: [7, 8, 9]

It is possible to provide indices that are out of bounds when slicing such as:

In [9]: l[:20] # 20 is out of index bounds, l[20] will raise an IndexError exception
Out[9]: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

In [11]: l[-20:] # -20 is out of index bounds, l[-20] will raise an IndexError exception
Out[11]: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

Keep in mind that the result of slicing a collection is a whole new collection. In addition, when using slice notation in assignments, the length of the slice assignment do not need to be the same. The values before and after the assigned slice will be kept, and the collection will shrink or grow to contain the new values:

In [16]: l[2:6] = list('abc') # assigning less elements than the ones contained in the sliced collection l[2:6]

In [17]: l
Out[17]: [0, 1, 'a', 'b', 'c', 6, 7, 8, 9]

In [18]: l[2:5] = list('hello') # assigning more elements than the ones contained in the sliced collection l [2:5]

In [19]: l
Out[19]: [0, 1, 'h', 'e', 'l', 'l', 'o', 6, 7, 8, 9]

If you omit the start and end index, you will make a copy of the collection:

In [14]: l_copy = l[:]

In [15]: l == l_copy and l is not l_copy
Out[15]: True

If the start and end indexes are omitted when performing an assignment operation, the entire content of the collection will be replaced with a copy of what is referenced:

In [20]: l[:] = list('hello...')

In [21]: l
Out[21]: ['h', 'e', 'l', 'l', 'o', '.', '.', '.']

Besides basic slicing, it is also possible to apply the following notation:

l[start:end:step]

where l is a collection, start is an inclusive index, end is an exclusive index, and step is a stride that can be used to take every nth item in l.

In [22]: l = list(range(10))

In [23]: l[::2] # take the elements which indexes are even
Out[23]: [0, 2, 4, 6, 8]

In [24]: l[1::2] # take the elements which indexes are odd
Out[24]: [1, 3, 5, 7, 9]

Using step provides a useful trick to reverse a collection in Python:

In [25]: l[::-1]
Out[25]: [9, 8, 7, 6, 5, 4, 3, 2, 1, 0]

It is also possible to use negative integers for step as the following example:

In[28]:  l[::-2]
Out[28]: [9, 7, 5, 3, 1]

However, using a negative value for step could become very confusing. Moreover, in order to be Pythonic, you should avoid using start, end, and step in a single slice. In case this is required, consider doing this in two assignments (one to slice, and the other to stride).

In [29]: l = l[::2] # this step is for striding

In [30]: l
Out[30]: [0, 2, 4, 6, 8]

In [31]: l = l[1:-1] # this step is for slicing

In [32]: l
Out[32]: [2, 4, 6]
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