Stackoverflow 社區：
我正在嘗試基于另一個串列的值的隨機采樣創建一個帶有回圈的子串列串列；并且每個子串列都具有沒有重復項或已添加到先前子串列中的值的限制。
假設（示例）我有一個主串列：
[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
#I get:
[[1,13],[4,1],[8,13]]
#I WANT:
[[1,13],[4,9],[8,14]] #(no duplicates when checking previous sublists)
我認為它會起作用的真正代碼如下（作為草稿）：
matrixvals=list(matrix.index.values) #list where values are obtained
lists=[[]for e in range(0,3)] #list of sublists that I want to feed
vls=[] #stores the values that have been added to prevent adding them again
for e in lists: #initiate main loop
for i in range(0,5): #each sublist will contain 5 different random samples
x=random.sample(matrixvals,1) #it doesn't matter if the samples are 1 or 2
if any(x) not in vls: #if the sample isn't in the evaluation list
vls.extend(x)
e.append(x)
else: #if it IS, then do a sample but without those already added values (line below)
x=random.sample([matrixvals[:].remove(x) for x in vls],1)
vls.extend(x)
e.append(x)
print(lists)
print(vls)
當我得到以下資訊時，它不起作用：
[[[25], [16], [15], [31], [17]], [[4], [2], [13], [42], [13]], [[11], [7], [13], [17], [25]]]
[25, 16, 15, 31, 17, 4, 2, 13, 42, 13, 11, 7, 13, 17, 25]
如您所見，數字 13 重復了 3 次，我不明白為什么
我想要：
[[[25], [16], [15], [31], [17]], [[4], [2], [13], [42], [70]], [[11], [7], [100], [18], [27]]]
[25, 16, 15, 31, 17, 4, 2, 13, 42, 70, 11, 7, 100, 18, 27] #no dups
另外，有沒有辦法將 sample.random 結果轉換為值而不是串列？（獲得）：
[[25,16,15,31,17]], [4, 2, 13, 42,70], [11, 7, 100, 18, 27]]
另外，實際上最終的結果不是子串列串列，實際上是字典（上面的代碼是解決字典問題的草稿），有沒有辦法在字典中獲取以前的方法？使用我現在的代碼，我得到了下一個結果：
{'1stkey': {'1stsubkey': {'list1': [41,
40,
22,
28,
26,
14,
41,
15,
40,
33],
'list2': [41, 40, 22, 28, 26, 14, 41, 15, 40, 33],
'list3': [41, 40, 22, 28, 26, 14, 41, 15, 40, 33]},
'2ndsubkey': {'list1': [21,
7,
31,
12,
8,
22,
27,...}
而不是那個結果，我想要以下內容：
{'1stkey': {'1stsubkey': {'list1': [41,40,22],
'list2': [28, 26, 14],
'list3': [41, 15, 40, 33]},
'2ndsubkey': {'list1': [21,7,31],
'list2':[12,8,22],
'list3':[27...,...}#and so on
有沒有辦法同時解決 list 和 dict 問題？任何幫助將不勝感激；即使只有串列問題，我也可以取得一些進展
謝謝大家
uj5u.com熱心網友回復：
我意識到您可能更感興趣的是找出您的特定方法不起作用的原因。但是，如果我了解您想要的行為，我可能會提供替代解決方案。發布我的答案后，我會看看你的嘗試。
random.sample讓您k從population（集合、串列等）中采樣專案數。如果集合中沒有重復的元素，則保證您的隨機樣本中沒有重復：
from random import sample
pool = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
num_samples = 4
print(sample(pool, k=num_samples))
可能的輸出：
[9, 11, 8, 7]
>>>
無論您運行此代碼段多少次，您的隨機樣本中都不會出現重復的元素。這是因為random.sample不會生成隨機物件，它只是隨機選擇集合中已經存在的專案。例如，這與從一副牌中隨機抽牌或抽獎號碼時所采用的方法相同。
在您的情況下，pool是從可能的唯一編號池中選擇您的樣本。您想要的輸出似乎是三個串列的串列，其中每個子串列中有兩個樣本。random.sample與其呼叫三次，每個子串列呼叫一次，不如呼叫一次k=num_sublists * num_samples_per_sublist：
from random import sample
pool = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
num_sublists = 3
samples_per_sublist = 2
num_samples = num_sublists * samples_per_sublist
assert num_samples <= len(pool)
print(sample(pool, k=num_samples))
可能的輸出：
[14, 10, 1, 8, 6, 3]
>>>
OK, so we have six samples rather than four. No sublists yet. Now you can simply chop this list of six samples up into three sublists of two samples each:
from random import sample
pool = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
num_sublists = 3
samples_per_sublist = 2
num_samples = num_sublists * samples_per_sublist
assert num_samples <= len(pool)
def pairwise(iterable):
yield from zip(*[iter(iterable)]*samples_per_sublist)
print(list(pairwise(sample(pool, num_samples))))
Possible output:
[(4, 11), (12, 13), (8, 15)]
>>>
Or if you really want sublists, rather than tuples:
def pairwise(iterable):
yield from map(list, zip(*[iter(iterable)]*samples_per_sublist))
EDIT - just realized that you don't actually want a list of lists, but a dictionary. Something more like this? Sorry I'm obsessed with generators, and this isn't really easy to read:
keys = ["1stkey"]
subkeys = ["1stsubkey", "2ndsubkey"]
num_lists_per_subkey = 3
num_samples_per_list = 5
num_samples = num_lists_per_subkey * num_samples_per_list
min_sample = 1
max_sample = 50
pool = list(range(min_sample, max_sample 1))
def generate_items():
def generate_sub_items():
from random import sample
samples = sample(pool, k=num_samples)
def generate_sub_sub_items():
def chunkwise(iterable, n=num_samples_per_list):
yield from map(list, zip(*[iter(iterable)]*n))
for list_num, chunk in enumerate(chunkwise(samples), start=1):
key = f"list{list_num}"
yield key, chunk
for subkey in subkeys:
yield subkey, dict(generate_sub_sub_items())
for key in keys:
yield key, dict(generate_sub_items())
print(dict(generate_items()))
Possible output:
{'1stkey': {'1stsubkey': {'list1': [43, 20, 4, 27, 2], 'list2': [49, 44, 18, 8, 37], 'list3': [19, 40, 9, 17, 6]}, '2ndsubkey': {'list1': [43, 20, 4, 27, 2], 'list2': [49, 44, 18, 8, 37], 'list3': [19, 40, 9, 17, 6]}}}
>>>
轉載請註明出處，本文鏈接：https://www.uj5u.com/qukuanlian/447816.html