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I'm trying to write a csv as follows: data[0][0] should be in row 1 column A, data[0][1] should be in row 1 column B, data[1][0] should be in row 2 column A, data[1][1] should be in row 2 column B and so on.

If I use something inefficient like the following code, I can write to the csv as intended.

                for i in range(0, len(data)-lg):
                    writer.writerow([data[i][0], data[i][1], data[i][2],...])

I really need something like the following code. But, this code writes everything in Column A. How can I modify it to accomplish my objective? Thanks.

            for i in range(0, len(data)-lg):
                for j in range(0, lg+6):
                    writer.writerow([data[i][j]])
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3 Answers 3

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I think the idiomatic way to do this would be as such (not fully understanding your lg variable):

num_cols = lg + 6
for row in data:
    writer.writerow(row[:num_cols])

If you weren't doing any processing on the rows in the loop, you could do it even more concisely:

writer.writerows(data)

Edit 1 Updated based on comment explaining lg variable

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7 Comments

Thanks. lg is a number between 10 and 30. len(data) can easily be 10,000+. In a sense, lg (actually, lg+6) is the number of columns that the csv will have. So, there will be 10,000+ rows and lg+6 columns. lg is important because it ties to other loops in the code. Before posting this question, I tried Dan D.'s solution, but the problem is that each row in data looks like this: ['blah', '11/12/2011', 6, 7, 8, 9]. So, when row is written to the csv, it's written in one cell with the brackets and single quotes. I need each value written in a separate column, which is why I tried data[i][j].
OK, so if lg is the number of columns you probably want to slice on the individual row and not the data. If you can confirm, I'll update the code. The csv module should take care of splitting your rows into the appropriate columns, but it sounds like you're saying it's not working for you. I've used the module numerous times so I think there's probably something in the code you haven't presented, perhaps in how the writer is initialized.
I'm using with open('test.csv', 'wb') as f: and writer = csv.writer(f)
Sorry if this is a dumb question or insulting your level of expertise, but what are you using to open your CSV file to verify its contents? Are you using a text editor or some full-blown spreadsheet software. I ask only because some spreadsheet software might have a different default dialect for CSV (e.g. semicolons instead of commas) and hence not interpreting it correctly.
Just for kicks, revert your code back to a version you'd expect to work and open the file up in a text editor like notepad/wordpad and see if the columns are delimited properly with commas. I'm wondering if you just need to set some options when you open the file (for example, Open Office Calc has a prompt that lets you specify the formatting of the CSV file).
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why not?

for row in data:
    writer.writerow(row)
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1 Comment

The reason I didn't go this way is because the row in data looks like this: ['blah', '11/12/2011', 6, 7, 8, 9]. So, each csv row populates in exactly that format all in one cell. The fix I came up with for this was specifying each item in the list. Do you know a better way to get blah in row1 colA, 11/12/2011 in row1 colB and so on?
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writerow() takes a row (sequence of strings or numbers) as input. Your second code passes only one element to writerow() so the element is written in separate row (i.e. each element in column A). In your first example you do the right thing by passing the entire collection of data row to writerow() so it works. As Dan D. has suggested a more pythonic way is

for x in data:
    writer.writerow(x)
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2 Comments

Thanks. That makes sense. See my comment to Dan D. above.
I did not fully understand it. Depending on what delimiter you have defined in csv, the data will be written to a row with each element of list separated by delimiter (default is ,)

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