Бинирование списка в python

Это на самом деле решает ряд проблем, а не только часть биннинга. Я включил код для анализа блоков в начале файла данных, чтобы вы могли получить все пиковые данные.

import numpy
import re
import matplotlib.pyplot as plt

f = open('sample_data.txt')
f.next()

pair = re.compile(r'# (.*?)[ \t]*:[ \t]*([0-9e\.-]+).*')

blocks = []
block = {}
blocks.append(block)

for line in f:
    if line[0] <> '#': 
        blocks.append(block)
        break
    line = line.strip()
    m = pair.match(line)
    if m:
        print line
        key, valstr = m.groups()
        print key, valstr
        try:
            value = float(valstr)
        except:
            value = valstr
        block[key] = value

    if (line == "#") and len(block) > 0:
        blocks.append(block)
        block = {}

peaks = sorted([block for block in blocks if 'PEAK' in block], 
               key=lambda b: b['PEAK'])
print peaks

colnames = ['HJD', 'Tamuz-corrected magnitude', 'Magnitude error']
data = numpy.loadtxt(f, [(colname, 'float64') for colname in colnames])

Nbins = 50
for peak in peaks:
    plt.figure()
    phase, _ = numpy.modf((data['HJD'] + 10*peak['Period (days)'] - peak['Epoch'])/peak['Period (days)'])
    mag = data['Tamuz-corrected magnitude']

    # use numpy.histogram to calculate the sum and the number of points in the bins
    sums, _ = numpy.histogram(phase, bins=Nbins, weights=mag)
    N, bin_edges = numpy.histogram(phase, bins=Nbins)

    # We'll plot the value at the center of each bin
    centers = (bin_edges[:-1] + bin_edges[1:])/2

    plt.scatter(phase, mag, alpha=0.2)
    plt.plot(centers, sums/N, color='red', linewidth=2)
plt.show()

Полный код: http://pastebin.com/4aBjZC7Q

Вот фрагмент, выполняющий биннинг:

x = []  # Column 1: HJD
y = []  # Column 2: Tamuz-corrected magnitude

# code to read "sample_data.txt" into lists x and y
# the full code in http://pastebin.com/4aBjZC7Q includes this part as well

import numpy as np

# these will characterize the bins
startBinValue = 5060
endBinValue = 5176
binSize = 0.1

# numpy.arange() will generate the bins of size binSize in between the limiting values
bins = np.arange(startBinValue, endBinValue, binSize)
# numpy.digitize() will "bin" the values; i.e. x_binned[i] will give the bin number of value in i-th index
x_binned = np.digitize(x, bins)
y_numpyArray = np.array(y)

# There is quite a bit of jugglery here.
# x_binned==i gives a boolean of size x_binned
# y_numpyArray[x_binned==i] returns the elements of y_numpyArray where the boolean is true
# The len() check is to make sure that mean() is not called for an empty array (which results in NAN
y_means = np.array([
    y_numpyArray[x_binned == i].mean()
    if len(y_numpyArray[x_binned == i]) > 0
    else 0
    for i in range(1, len(bins))])

# binnedData is a list of tuples; tuple elements are bin's-low-limit, bin's-high-limit, y-mean value
binnedData = [(bins[i], bins[i + 1], y_means[i]) for i in range(len(y_means))]

Я сильно прокомментировал код. Но чтобы понять всю функциональность функций numpy, обратитесь к numpy .digitize(), numpy.arange()< /а>. Я использовал numpy.arange() для создания корзин. предполагая, что вы заранее знаете размеры бинов, но если вам нужно фиксированное количество бинов (скажем, 100 бинов для x-данных), используйте numpy.linspace().