编写CSV文件python

Question

So i have a list:

>>> print references
>>> ['Reference xxx-xxx-xxx-007 ', 'Reference xxx-xxx-xxx-001 ', 'Reference xxx-xxx-xxxx-00-398 ', 'Reference xxx-xxx-xxxx-00-399']

(The list is much longer than that)

I need to write a CSV file wich would look this:

Column 1:
Reference xxx-xxx-xxx-007
Reference xxx-xxx-xxx-001
[...]

I tried this :

c = csv.writer(open("file.csv", 'w'))
for item in references:
    c.writerows(item)
Or:
for i in range(0,len(references)):
    c.writerow(references[i])

But when I open the csv file created, I get a window asking me to choose the delimiter No matter what, I have something like R,e,f,e,r,e,n,c,es

Answer 1

writerows takes a sequence of rows, each of which is a sequence of columns, and writes them out.

But you only have a single list of values. So, you want:

for item in references:
    c.writerow([item])

Or, if you want a one-liner:

c.writerows([item] for item in references)

The point is, each row has to be a sequence; as it is, each row is just a single string.

So, why are you getting R,e,f,e,r,e,n,c,e,… instead of an error? Well, a string is a sequence of characters (each of which is itself a string). So, if you try to treat "Reference" as a sequence, it's the same as ['R', 'e', 'f', 'e', 'r', 'e', 'n', 'c', 'e'] .

---

In a comment, you asked:

Now what if I want to write something in the second column ?

Well, then each row has to be a list of two items. For example, let's say you had this:

references = ['Reference xxx-xxx-xxx-007 ', 'Reference xxx-xxx-xxx-001 ']
descriptions = ['shiny thingy', 'dull thingy']

You could do this:

csv.writerows(zip(references, descriptions))

Or, if you had this:

references = ['Reference xxx-xxx-xxx-007 ', 'Reference xxx-xxx-xxx-001 ', 'Reference xxx-xxx-xxx-001 ']
descriptions = {'Reference xxx-xxx-xxx-007 ': 'shiny thingy', 
                'Reference xxx-xxx-xxx-001 ': 'dull thingy']}

You could do this:

csv.writerows((reference, descriptions[reference]) for reference in references)

The key is, find a way to create that list of list s—if you can't figure it out all in your head, you can print all the intermediate steps to see what they look like—and then you can call writerows . If you can only figure out how to create each single row one at a time, use a loop and call writerow on each row.

---

But what if you get the first column values, and then later get the second column values?

Well, you can't add a column to a CSV; you can only write by row, not column by column. But there are a few ways around that.

First, you can just write the table in transposed order:

c.writerow(references)
c.writerow(descriptions)

Then, after you import it into Excel, just transpose it back.

Second, instead of writing the values as you get them, gather them up into a list, and write everything at the end. Something like this:

rows=[[item] for item in references] 
# now rows is a 1-column table
# ... later
for i, description in enumerate(descriptions):
    values[i].append(description)
# and now rows is a 2-column table
c.writerows(rows)

If worst comes to worst, you can always write the CSV, then read it back and write a new one to add the column:

with open('temp.csv', 'w') as temp:
    writer=csv.writer(temp)
    # write out the references
# later
with open('temp.csv') as temp, open('real.csv', 'w') as f:
    reader=csv.reader(temp)
    writer=csv.writer(f)
    writer.writerows(row + [description] for (row, description) in zip(reader, descriptions))

Answer 2

writerow writes the elements of an iterable in different columns. This means that if your provide a tuple, each element will go in one column. If you provide a String, each letter will go in one column. If you want all the content in the same column do the following:

c = csv.writer(open("file.csv", 'wb'))
c.writerows(references)

or

for item in references:
    c.writerow(references)

Answer 3

c = csv.writer(open("file.csv", 'w'))
c.writerows(["Reference"])

# cat file.csv
R,e,f,e,r,e,n,c,e

but

c = csv.writer(open("file.csv", 'w'))
c.writerow(["Reference"])


# cat file.csv
Reference

Would work as others have said.

My original answer was flawed due to confusing writerow and writerows .