Nested List - Strings to Float

Question

I have nested list of the following form:

my_list = [['Some1', '2', '3.6', '4.5', 'GB2', '6'],
           ['Some2', '3.9', '4', '5', 'HG5', '7.3'],
           ['Some3', '4', '5', '6.1', 'H2D', '8.9']]

Every element of each sublist is a string, but I'd like to turn all purely numeric strings into floats.

So I try the following code:

for row in my_list:
    for k, item in enumerate(row):
        if k in (1, 2, 3, 5):
            item = float(item)

Unfortunately, the nested list remains unchanged. I'm sure I'm making a simple mistake, but I can't see it. Any help (and context) would be appreciated.

Sven Marnach · Accepted Answer · 2012-03-19 18:02:06Z

4

Reassigning the name item will never change the list. Try this instead:

for row in my_list:
    for k in (1, 2, 3, 5):
        row[k] = float(row[k])

Note that Python names are just tags attached to some objects. A line like

item = float(item)

will compute float(item) and then bind the name item to the returned object. This does not do anything to the object item pointed to before. Moreover, in this case it is impossible to mutate the latter object in place -- strings are immutable.

answered Mar 19, 2012 at 18:02

Sven Marnach

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2 Comments

donopj2 Over a year ago

Which would be considered more Pythonic in this case, a for loop or a list comprehension creating my_list2 like DSM suggests? Is one or the other preferred? My actual lists can reach 100,000 + rows.

Sven Marnach Over a year ago

@PatrickD: The two answers do different things. The code in my answer mutates the original list in place. The code in DSM's answer creates a new list. Another difference is that the code in my answer hardcodes which columns to convert, and will error out if some value cannot be converted. The code in DSM's answer tries to convert every item, and uses the floating point value instead of the original value if the conversion succeeds. The code in this answer is faster and consumes less memory, provided you want these semantics.

DSM · Accepted Answer · 2012-03-19 18:02:23Z

4

You're not affecting the original list; you're simply making a float from some of the elements and giving the name "item" to it in turn. You could use something like

row[k] = float(item)

but I'd probably do something like

def floatify(x):
    try:
        return float(x)
    except ValueError:
        return x

my_list2 = [[floatify(x) for x in row] for row in my_list]

answered Mar 19, 2012 at 18:02

DSM

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Comments

fraxel · Accepted Answer · 2012-03-19 18:08:34Z

1

for index1, row in enumerate(my_list):
    for index2, item in enumerate(row):
        try:
            my_list[index1][index2] = (float(item))
        except ValueError:
            pass
print my_list
[['Some1', 2.0, 3.6000000000000001, 4.5, 'GB2', 6.0], 
['Some2', 3.8999999999999999, 4.0, 5.0, 'HG5', 7.2999999999999998], 
['Some3', 4.0, 5.0, 6.0999999999999996, 'H2D', 8.9000000000000004]]

answered Mar 19, 2012 at 18:08

fraxel

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Comments

Anurag Uniyal · Accepted Answer · 2012-03-19 18:02:28Z

0

You will need to reassign row elements to new calculated value e.g.

row[k] = float(item)

answered Mar 19, 2012 at 18:02

Anurag Uniyal

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