Openpyxl How to get row from worksheet by index

Question

Using Openpyxl and python3.5, I tried getting the first row from an excel worksheet using a subscript but I an error.

# after getting filename
# after loading worksheet
# to get the first row of the worksheet
first_row = worksheet.rows[0]

# I get 
Traceback (most recent call last):
      File "<pyshell#54>", line 1, in <module>
      first_row = phc_th_sheet.rows[1]
TypeError: 'generator' object is not subscriptable

In relation to getting the first row, I've also tried first_row = worksheet.(row=1) # and first_row = worksheet.rows[:1]

None worked. Any suggestions or is the feature not available in openpyxl? I've been to the documentation at https://openpyxl.readthedocs.io/en/default/ but I found nothing helpful enough to index and select rows

Solution 1

I finally found the answer in the documentation:

first_row = worksheet[1]
# worksheet[row_index_from_1]

This worked for me.

Solution 2

The error TypeError: 'generator' object is not subscriptable. Means that you are trying to access by index a generator, which doesn't have one, because it creates the elements as you iterate through it.

You can solve it easily, cast it to a list to get the element you want:

first_row = list(worksheet.rows)[0]

or iterate thought the rows:

for row in worksheet.rows:
    foo(row)

This is because, even if both are iterables, lists and generators can behave quite differently, you can get it better explained here:

https://wiki.python.org/moin/Generators

https://docs.python.org/3/library/stdtypes.html#iterator-types

https://docs.python.org/3/library/stdtypes.html#sequence-types-list-tuple-range

Comments

[Charlie Clark]

This is covered in the documentation.

[Charlie Clark]

Yes, we converted ws.rows into a generator so that behaviour across different implementations would be consistent and also to encourage this syntax for ad-hoc queries. Programmatic access is best done using ws.iter_rows() and, where possible, ws.iter_cols().

[Ben Liyanage]

This is now worksheet.rows[1]. Going to update the answer.

[Ben Liyanage]

This is not a great answer. It's better to just get the row directly, rather then explicitly converting the iterator to a list.

[Manuel Alvarez]

Sure it is. However, as I usually manipulate excel spreadsheets with pandas and I'm not familiar with the internals of openpyxl, just tried to give an explanation of the error and how to get a workaround.

[ebt]

@BenLiyanage the issue is you cant because the method is now a generator. the list method emulates accessing the row directly. Note that it pulls all the data into memory first. like DOM and StaX considerations, if you have a large dataset you'll want to iterate using the second example.

[Ben Liyanage]

Yup, I fixed the other answer. =D Happy bug hunting guys!

[Erdem KAYA]

I understand @ebt's explanation generators, but I still can't get how one could access to the n_th row with the "second example" presented in the answer.

[ebt]

@ErdemKAYA I would suggest using the enumerate method. docs.python.org/3/library/functions.html#enumerate you can use an if statement to identify which row to take action on.

[hsc]

@CharlieClark after the rows variable is a generator, do we still have a way to access row by index now?

[Mark Dickinson]

@BenLiyanage: I've rolled back your edit. I suspect you were using an older version of openpyxl. worksheet.rows is a generator, so worksheet.rows[1] won't work.

[Ben Liyanage]

If version is important, can you please specify it in the answer mark? Thanks for keeping the answer fresh! I don’t have access to this codebase any more so I do not know what version I was using.

[Flying Thunder]

this no longer works in the newest version