我有一本字典,其中有一个列表,该列表具有成对的数字作为子列表, 我正在尝试检查子列表之间是否有重叠。
{'5ykw.pdb': [[10, 22], [33, 40], [39, 51], [63, 71], [94, 105]]}
[33,40]和[39,51]之间有重叠,因此我想将它们合并为:
{'5ykw.pdb': [[10, 22], [33, 51], [63, 71], [94, 105]]}
答案 0 :(得分:0)
您可以将reduce
与自定义merge
函数一起使用来创建新列表:
from functools import reduce
def merge(acc, curr):
if not len(acc) or acc[-1][1] < curr[0]:
acc.append(curr)
return acc
acc[-1][1] = curr[1] # update last element in accumulator
return acc
data = {'5ykw.pdb': [[10, 22], [33, 40], [39, 51], [63, 71], [94, 105]]}
data['5ykw.pdb'] = reduce(merge, data['5ykw.pdb'], [])
print(data)
# {'5ykw.pdb': [[10, 22], [33, 51], [63, 71], [94, 105]]}
答案 1 :(得分:0)
您可以使用此功能简单地“消除重叠”列表:
def deoverlap(lst):
if not lst:
return []
lst = [sorted(pair) for pair in lst] # sort pairs (leave out if not needed)
lst = sorted(lst) # sort by first item (breaking ties by second item)
out = []
prev = lst[0]
for pair in lst[1:]:
if prev[1] >= pair[0]:
if prev[1] < pair[1]:
prev[1] = pair[1]
else:
out.append(prev)
prev = pair
out.append(prev)
return out
dct = {'5ykw.pdb': [[10, 22], [33, 40], [39, 51], [63, 71], [94, 105]]}
dct['5ykw.pdb'] = deoverlap(dct['5ykw.pdb'])
print(dct) # prints {'5ykw.pdb': [[10, 22], [33, 51], [63, 71], [94, 105]]}
这里唯一的假设是deoverlap()
的输入是可比较类型(通常是数字)的成对列表,其中每对都是长度为2的列表。
对在内部进行排序,然后按第一项进行排序,如果前一对的最大值≥当前对的最小值,则将其合并。如果在相等时不应该合并,则deoverlap()
的第9 行应变为
if prev[1] > pair[0]:
答案 2 :(得分:0)
您可以使用广度优先搜索的递归形式:
def overlap(a, b) -> bool:
return a[-1] >= b[0] and a[-1] < b[-1]
def group(d, _c, seen):
return [_c,
[i if i not in seen else group(d, i, seen+[i]) for i in d if overlap(_c, i)]]
r = {'5ykw.pdb': [[10, 22], [33, 40], [39, 51], [63, 71], [94, 105]]}
new_data = [group(r['5ykw.pdb'], i, []) for i in r['5ykw.pdb'] if not any(overlap(c, i) for c in r['5ykw.pdb'])]
final_data = [a if not b else [a[0], max(h for _, h in b)] for a, b in new_data]
输出:
[[10, 22], [33, 51], [63, 71], [94, 105]]
这也适用于具有大量重叠的输入:
r = {'5ykw.pdb':[[15, 20], [18, 21], [19, 30]]}
new_data = [group(r['5ykw.pdb'], i, []) for i in r['5ykw.pdb'] if not any(overlap(c, i) for c in r['5ykw.pdb'])]
final_data = [a if not b else [a[0], max(h for _, h in b)] for a, b in new_data]
输出:
[[15, 30]]