我正在寻找一种方法,将包含数百个条目的数据集放入 20 个容器中。但不使用 pandas(剪切)和 numpy(数字化)等大模块。 有人能想到比 18 个 elif 更好的解决方案吗?
最佳答案
您需要做的就是找出每个元素位于哪个容器中。考虑到容器的大小(如果它们是统一的),这相当简单。从数组中,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
,该元素将落入 bin 编号 int((elem - minval)/binwidth)
。这就留下了 elem == maxval
的边缘情况。在这种情况下,bin编号等于nbins(nbins +第1个bin,因为python是从零开始的),因此我们必须减少bin编号仅此一例。
因此我们可以编写一个函数来执行此操作:
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)
这给出
[(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)]
这看起来就像我们所期望的。
对于非均匀分箱,不再是算术计算。在本例中,元素 elem
位于下限小于 elem
且上限大于 elem
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
关于python-3.x - 无需 pandas/numpy 即可对数据进行 Pythonic 方式,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/64995641/