Extract Floating Point Numbers from Array

Question

I have an array:

[(0, b'C', 5.8816, 6.0, 0.1184, 4.2631, 4.2631, 0.0)
 (1, b'H', 0.8495, 1.0, 0.1505, 0.951, 0.951, 0.0)
 (2, b'Br', 35.0064, 35.0, -0.0064, 1.2192, 1.2192, -0.0)
 (3, b'Cl', 17.0401, 17.0, -0.0401, 1.2405, 1.2405, -0.0)
 (4, b'F', 9.2225, 9.0, -0.2225, 1.0449, 1.0449, -0.0)]

from which I want to extract single columns of floating point numbers fro use further on in the program.

When I try the following:

QA = []
idx_IN_columns = [5]
QA = data[idx_IN_columns]

I get:

Traceback (most recent call last):

  File "<ipython-input-22-4e6a1b6a3f36>", line 1, in <module>
    runfile('C:/Users/Steve/Python/Testing/ReadFile_mpa_1.py', wdir='C:/Users/Steve/Python/Testing')

  File "C:\Anaconda3\lib\site-packages\spyder\utils\site\sitecustomize.py", line 866, in runfile
    execfile(filename, namespace)

  File "C:\Anaconda3\lib\site-packages\spyder\utils\site\sitecustomize.py", line 102, in execfile
    exec(compile(f.read(), filename, 'exec'), namespace)
    enter code here

  File "C:/Users/Steve/Python/Testing/ReadFile_mpa_1.py", line 34, in <module>
    QA = data[idx_IN_columns]

IndexError: index 5 is out of bounds for axis 1 with size 5

Assistance will be much appreciated.

Thanks in advance

are you trying to get the element with index 5 from the array? — Junbang Huang
– Junbang Huang, Commented May 7, 2017 at 15:37
the array is confusing, is it a 2d array of just 1d array with tuple as its element. are you mising the commas? — Junbang Huang
– Junbang Huang, Commented May 7, 2017 at 15:38
@Steve Note that your full name shows in the paths of the error message you posted. — handle
– handle, Commented May 7, 2017 at 15:54

handle · Accepted Answer · 2017-05-07 16:09:55Z

1

You cannot access a list element with a list as index (see Sequence Types), you need to iterate. Your data sets are Tuples; they are similar to Lists, but cannot be modified. By the way, your data is malformed (lacking commas):

#!python3
# coding=utf-8

l = [
    (0, b'C', 5.8816, 6.0, 0.1184, 4.2631, 4.2631, 0.0),
    (1, b'H', 0.8495, 1.0, 0.1505, 0.951, 0.951, 0.0),
    (2, b'Br', 35.0064, 35.0, -0.0064, 1.2192, 1.2192, -0.0),
    (3, b'Cl', 17.0401, 17.0, -0.0401, 1.2405, 1.2405, -0.0),
    (4, b'F', 9.2225, 9.0, -0.2225, 1.0449, 1.0449, -0.0)
    ]

print( [e[2] for e in l] ) # list comprehension
# [5.8816, 0.8495, 35.0064, 17.0401, 9.2225]

# equivalent loop
result = []
for e in l:
    result.append(e[2])
print(result)

A little more advanced, as generator function:

# iterator
def col(l, n=0):
    for e in l:
        yield e[n]

print(list(col(l,2)))

for data in col(l,2):
    print(data)

edited May 7, 2017 at 16:09

answered May 7, 2017 at 15:51

handle

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Comments

Martin Valgur · Accepted Answer · 2017-05-07 16:07:01Z

0

Your data and what you are trying to do with it sounds like a good fit for Pandas, a library for working with data in a tabular format.

>>> import pandas as pd
>>> data = [(0, b'C', 5.8816, 6.0, 0.1184, 4.2631, 4.2631, 0.0),
 (1, b'H', 0.8495, 1.0, 0.1505, 0.951, 0.951, 0.0),
 (2, b'Br', 35.0064, 35.0, -0.0064, 1.2192, 1.2192, -0.0),
 (3, b'Cl', 17.0401, 17.0, -0.0401, 1.2405, 1.2405, -0.0),
 (4, b'F', 9.2225, 9.0, -0.2225, 1.0449, 1.0449, -0.0)]
>>> df = pd.DataFrame(data)
>>> df
   0      1        2     3       4       5       6    7
0  0   b'C'   5.8816   6.0  0.1184  4.2631  4.2631  0.0
1  1   b'H'   0.8495   1.0  0.1505  0.9510  0.9510  0.0
2  2  b'Br'  35.0064  35.0 -0.0064  1.2192  1.2192 -0.0
3  3  b'Cl'  17.0401  17.0 -0.0401  1.2405  1.2405 -0.0
4  4   b'F'   9.2225   9.0 -0.2225  1.0449  1.0449 -0.0

Then, to get column 5, just do df[5].

>>> df[5]
0    4.2631
1    0.9510
2    1.2192
3    1.2405
4    1.0449
Name: 5, dtype: float64

By using Pandas, you can also easily read from (pd.read_csv()) and write the data (df.to_csv()) to disk, for example.

answered May 7, 2017 at 16:07

Martin Valgur

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1 Comment

Steve Over a year ago

I had to make a slight modification to you solution.

arboreal84 · Accepted Answer · 2017-05-07 15:50:22Z

0

data = [(0, b'C', 5.8816, 6.0, 0.1184, 4.2631, 4.2631, 0.0),
 (1, b'H', 0.8495, 1.0, 0.1505, 0.951, 0.951, 0.0),
 (2, b'Br', 35.0064, 35.0, -0.0064, 1.2192, 1.2192, -0.0),
 (3, b'Cl', 17.0401, 17.0, -0.0401, 1.2405, 1.2405, -0.0),
 (4, b'F', 9.2225, 9.0, -0.2225, 1.0449, 1.0449, -0.0)]

numbers = [n[2:] for n in data]

Result

[(5.8816, 6.0, 0.1184, 4.2631, 4.2631, 0.0), (0.8495, 1.0, 0.1505, 0.951, 0.951, 0.0), (35.0064, 35.0, -0.0064, 1.2192, 1.2192, -0.0), (17.0401, 17.0, -0.0401, 1.2405, 1.2405, -0.0), (9.2225, 9.0, -0.2225, 1.0449, 1.0449, -0.0)]

edited May 7, 2017 at 15:50

answered May 7, 2017 at 15:44

arboreal84

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Comments

Ben Schmidt · Accepted Answer · 2017-05-07 15:53:05Z

0

Assuming your array is meant to have commas at the end, like

data = [(0, b'C', 5.8816, 6.0, 0.1184, 4.2631, 4.2631, 0.0),
 (1, b'H', 0.8495, 1.0, 0.1505, 0.951, 0.951, 0.0),
 (2, b'Br', 35.0064, 35.0, -0.0064, 1.2192, 1.2192, -0.0),
 (3, b'Cl', 17.0401, 17.0, -0.0401, 1.2405, 1.2405, -0.0),
 (4, b'F', 9.2225, 9.0, -0.2225, 1.0449, 1.0449, -0.0)]

Which is actually a list of tuples (not really an array), you could do

QA = [row[5] for row in data]

or

from operator import itemgetter

QA = map(itemgetter(5), data)

answered May 7, 2017 at 15:53

Ben Schmidt

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