Save list to csv with two decimal digits in python?

Question

I have data which looks like

[('A',5.3748),('B',8.324203),('C',3.492)]

I'm saving it to a csv with the following lines:

with open('output.csv','wb') as fp:
    a = csv.writer(fp,delimiter = ",")
    a.writerows(data)

Works like a charm except that I would like to display my values with only two decimal digits. Earlier in my code I zip two lists to get the data. So I could make the change before if need be.

Thanks!

You could turn all your floats into formatted strings, those should save correctly. — thegrinner, Oct 07 '13 at 14:21
And can I do this for a whole list in one go or do I need a loop? — marc, Oct 07 '13 at 14:25
Depends what you mean by "*display my values*"... If you're opening the file into Excel or similar then you're probably better off using the tools available there to format the display of the data rather than truncating the data on output... Otherwise - string formatting is the correct approach — Jon Clements, Oct 07 '13 at 14:25

score 6 · Answer 1 · answered Oct 07 '13 at 14:29

You can use the string formatting syntax:

with open('output.csv','wb') as fp:
    a = csv.writer(fp,delimiter = ",")
    a.writerows(map(lambda t: (t[0], "%.2f" % t[1]), data))

The trick is here:

>> "%.2f" % a_number

will print the number with two decimal digits, while

map(lambda t: (t[0], "%.2f" % t[1]), data)

will apply this transformation to every tuples in the list data. (see http://docs.python.org/2/library/functions.html#map)

score 2 · Accepted Answer · answered Oct 07 '13 at 14:24

2

If you can isolate the number you want to round, you can format it to two decimal places and store it as a string like this:

"%.2f" % myNum

answered Oct 07 '13 at 14:24

Bucket

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score 1 · Answer 3 · answered Oct 07 '13 at 14:25

1

from itertools import imap
with open('output.csv','wb') as fp:
  a = csv.writer(fp,delimiter = ",")
  a.writerows(imap(lambda x: (x[0], round(x[1], 2)), data))

answered Oct 07 '13 at 14:25

Maciej Gol

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thegrinner · Answer 4 · 2013-10-07T14:31:06.083

Try formatting your decimals using str.format():

temp = [('A',5.3748),('B',8.324203),('C',3.492)]
newlist = []
for a, b in temp:
    newlist.append((a, "{0:.2f}".format(b)))

As a list comprehension, you get:

data = [('A',5.3748),('B',8.324203),('C',3.492)]
data = [(a, "{0:.2f}".format(b)) for a, b in temp]

As @JonClements noted, you can replace "{0:.2f}".format(b) with format(b, '.2f'). In this case, that's probably the more readable approach.

score 1 · Answer 5 · answered May 11 '20 at 01:18

You can also use python's round() method.

round(5.3748, 2)

The advantage of using this with a csv writer is that it maintains it's number type instead of being converted to a string. This affects options like quoting non-numericals in the csv file.

If you don't need to worry about such formatting options I personally prefer formatting it to a string, since it aligns the correct number of decimal places even if they are zero. I.e. 5.3748 to "5.3700" instead of 5.37

score 0 · Answer 6 · answered Oct 07 '13 at 14:31

0

You can format the list elements as follow

linux:~ # python
>>> l=[('A',5.3748),('B',8.324203),('C',3.492)]
>>> l=[ (x[0], "%.2f" % x[1]) for x in l]
>>> l
[('A', '5.37'), ('B', '8.32'), ('C', '3.49')]

answered Oct 07 '13 at 14:31

LMC

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Save list to csv with two decimal digits in python?

6 Answers6