converting string to float (and removing decimal point) in python

Question

I'm trying to create a function that imports a csv file which holds records for population per year (as strings). It imports a file which has the year in the 3rd column and the population count in the 4th.

It should remove the decimal point '.' and display the resulting population.

16122.83
16223.248

should become

1612283
16223248

When I try to do this I get: in print_population_list year, population = row[2], float(row[3]) ValueError: could not convert string to float: POP. This is my code:

import csv
file = csv.reader(open(filename))

year, population = 0, 0

for row in file:
    year, population = row[2], float(row[3])
    print year,":", population,

To do this I figured it should first be converted to a float and be multiplied by the highest number of decimal places, after which all zero's at the end should be removed (since the data doesn't all have the same number of decimal places). But I'm stuck at the float conversion.

Roger that, I removed my comment.

ErlVolton
– ErlVolton

2014-10-17 17:46:43 +00:00
Commented Oct 17, 2014 at 17:46 — ErlVolton
– ErlVolton, Commented Oct 17, 2014 at 17:46

Jonathan Eunice · Accepted Answer · 2014-10-17 15:16:08Z

2

Most direct route:

>>> s = '16122.83'
>>> int(s.replace('.', ''))
1612283

While performance is probably not a big concern in your use case, a replace strategy is about 30% faster than the split-join strategy, based on a simple benchmark.

Benchmark Report
================

Options
-------

   name | rank | runs |     mean |        sd | timesBaseline
--------|------|------|----------|-----------|--------------
replace |    1 | 1000 | 0.009488 | 0.0006711 |           1.0
   join |    2 | 1000 |  0.01258 | 0.0007729 | 1.32589602108

Each of the above 2000 runs were run in random, non-consecutive order by
`benchmark` v0.1.5 (http://jspi.es/benchmark)

For this problem, int is what you seem to need. But for related problems, using float instead of int would keep you in the floating point realm. The round(value, places) function also might be handy.

edited Oct 17, 2014 at 15:16

answered Oct 17, 2014 at 15:01

Jonathan Eunice

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Comments

Cory Kramer · Accepted Answer · 2014-10-17 14:52:03Z

1

This should do the trick

def decToNum(s):
    return int(''.join(s.split('.')))

>>> s = '16122.83'
>>> decToNum(s)
1612283

answered Oct 17, 2014 at 14:52

Cory Kramer

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1 Comment

Roger That Over a year ago

this gives me: in decToNum return int(''.join(s.split('.'))) ValueError: invalid literal for int() with base 10: 'POP'

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