我正在處理一個有趣的 python 問題：給定一個整數(shù)、細菌和一個正整數(shù)常數(shù) K 的列表，如果有兩個元素 i 和 j 滿足以下條件：j < i <= j + K設計一個函數(shù)，使 i 留下并移除 j，并返回列表中元素的最小可能數(shù)量。功能： micro_world(bacteria, K)例如，bacteria = [101, 53, 42, 102, 101, 55, 54]K = 1最終結果應該是 [42,102,55] 因此應返回 3。相似地micro_world([101, 53, 42, 102, 101, 55, 54], 1) 會給我一個結果 3micro_world([20, 15, 10, 15, 20, 25], 5) 將給我一個結果 1我正在考慮使用列表理解來過濾掉不符合上述標準的元素，從而獲得我想要的元素。但我不確定如何繼續(xù)，因為它涉及列表中每個元素之間的關系。result = [ i for i in bacteria if ... ]我的 if 語句應該使用什么？如果這不是一個好方法，我該怎么辦？謝謝。編輯：這兩個答案都為我提供了非常好的指導。但是關于細菌列表中的重復值只有一件小事。例如，如果輸入bacteria = [3, 3]即使這只是一個塊，結果應該是 2，因為 3 ！> 3 因此兩者都不會被刪除。
2 回答
慕婉清6462132
TA貢獻1804條經(jīng)驗 獲得超2個贊
您實際上是在嘗試將您的數(shù)字列表分組到塊中，其中每個k數(shù)字與該塊中另一個數(shù)字的距離小于。因為我不擅長解釋事情，讓我們看一個例子：
bacteria = [101, 53, 42, 102, 101, 55, 54]
K = 1
首先，我們要對該列表進行排序，以便數(shù)字按大小排列：
[102, 101, 101, 55, 54, 53, 42]
現(xiàn)在我們遍歷列表并在每次較大的細菌無法吞下較小的細菌時創(chuàng)建一個新的數(shù)字塊：
[[102, 101, 101], [55, 54, 53], [42]]
最后我們計算chunk的數(shù)量，從而得到想要的結果：3。
代碼：
def micro_world(bacteria, k):
# sort the list descendingly
bacteria = sorted(bacteria, reverse=True)
# loop over the list but skip all the "swallowed" bacteria
i = 0
result = 0
while i < len(bacteria):
bacterium_size = bacteria[i]
# while the difference between the numbers is <= k, the smaller
# bacterium is swallowed
bigger_bacterium_exists = False
while i+1 < len(bacteria):
difference = bacterium_size - bacteria[i+1]
# if the difference is too big, it can't be swallowed
if difference > k:
break
# if there are two bacteria of the same size, they can't swallow
# each other. But if a bigger bacterium exists, it can swallow
# them both
if difference == 0 and not bigger_bacterium_exists:
break
# all conditions are met, the bacterium is swallowed
bacterium_size = bacteria[i+1]
i += 1
bigger_bacterium_exists = True
# at this point, the bacterium has swallowed everything it can.
# Increment the result counter and move on to the next bacterium.
result += 1
i += 1
return result
鳳凰求蠱
TA貢獻1825條經(jīng)驗 獲得超4個贊
這是一個使用的解決方案numpy：
import numpy as np
def micro_world(bacteria, K):
# convert bacteria list to a numpy array:
bacteria = np.asarray(bacteria)
# sort array and remember the indices used for sorting:
sarg = np.argsort(bacteria)
sortedbac = bacteria[sarg]
# compute differences between adjacent elements:
diff = np.ediff1d(sortedbac, K + 1)
# throw away elements that are too close to neighbors
# (find indices of the elements to keep):
idx = np.flatnonzero(diff > K)
# create a new list that meets selection criteria:
return bacteria[np.sort(sarg[idx])]
這是一個“純”的 Python 實現(xiàn)：
def micro_world(bacteria, K):
# sort array and remember the indices used for sorting:
sarg = [i[0] for i in sorted(enumerate(bacteria), key=lambda x: x[1])]
sortedbac = [bacteria[i] for i in sarg]
# compute differences between adjacent elements:
diff = [j - i for i, j in zip(sortedbac[:-1], sortedbac[1:])] + [K + 1]
# throw away elements that are too close to neighbors
# (find indices of the elements to keep):
idx = [i for i, v in enumerate(diff) if v > K]
# create a new list that meets selection criteria:
return [bacteria[i] for i in sorted([sarg[i] for i in idx])]
如果你只對元素的數(shù)量感興趣，而不對元素本身感興趣，那么你可以修改 ie 第二個版本，如下所示：
def micro_world(bacteria, K):
sortedbac = sorted(bacteria)
diff = [j - i for i, j in zip(sortedbac[:-1], sortedbac[1:])] + [K + 1]
return sum(1 for v in diff if v > K)
在numpy隨后的版本將成為：
def micro_world(bacteria, K):
return np.sum(np.ediff1d(np.sort(bacteria), K + 1) > K)
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