在没有pandas/numpy的情况下对数据进行装箱的Python方式

Question

我正在寻找一种方法，将数百个条目的数据集放入20个bin中。但是没有使用像pandas(剪切(和numpy(数字化(这样的大模块。有人能想出比18个elif更好的解决方案吗？

Answer 1

你所需要做的就是弄清楚每个元素在哪个bin中。考虑到bin的大小，如果它们是统一的，这是相当微不足道的。从您的数组中，您可以找到minval和maxval。然后，binwidth = (maxval - minval) / nbins。对于数组elem的元素，以及已知的最小值minval和bin宽度binwidth，该元素将属于bin编号int((elem - minval) / binwidth)。这就留下了边缘情况，其中CCD_。在这种情况下，bin数等于nbins(第nbins + 1个bin，因为python是基于零的(，所以我们必须为这一种情况递减bin数。

因此，我们可以编写一个函数来实现这一点：

import random
def splitIntoBins(arr, nbins, minval=None, maxval=None):
minval = min(arr) if minval is None else minval # Select minval if specified, otherwise min of data
maxval = max(arr) if maxval is None else maxval # Same for maxval

binwidth = (maxval - minval) / nbins # Bin width
allbins = [[] for _ in range(nbins)] # Pre-make a list-of-lists to hold values
for elem in arr:
binnum = int((elem - minval) // binwidth) # Find which bin this element belongs in
binindex = min(nbins-1, binnum) # To handle the case of elem == maxval
allbins[binindex].append(elem) # Add this element to the bin
return allbins
# Make 1000 random numbers between 0 and 1
x = [random.random() for _ in range(1000)]
# split into 10 bins from 0 to 1, i.e. a bin every 0.1
b = splitIntoBins(x, 10, 0, 1)
# Get min, max, count for each bin
counts = [(min(v), max(v), len(v)) for v in b]
print(counts)

这提供

[(0.00017731201786974626, 0.09983758434153, 101),
(0.10111204267013452, 0.19959594179848794, 97),
(0.20089309189822557, 0.2990120768922335, 100),
(0.3013915797055913, 0.39922131591077614, 90),
(0.4009006835799309, 0.49969892298935836, 83),
(0.501675740585966, 0.5999729295882031, 119),
(0.6010149249108184, 0.7000366124696699, 120),
(0.7008002068562794, 0.7970568220766774, 91),
(0.8018697850229161, 0.8990963218226316, 99),
(0.9000732426223624, 0.9967964437788829, 100)]

这看起来像我们所期望的。

对于非均匀bin，它不再是算术计算。在这种情况下，元素elem在具有小于elem的下界和大于elem的上界的bin中。

def splitIntoBins2(arr, bins):
binends = bins[1:]
binstarts = bins[:-1]
allbins = [[] for _ in binends] # Pre-make a list-of-lists to hold values
for elem in arr:
for i, (lower_bound, upper_bound) in enumerate(zip(binstarts, binends)):
if upper_bound >= elem and lower_bound <= elem:
allbins[i].append(elem) # Add this element to the bin
break
return allbins