Question

以下代码获取匹配值的索引：

x = ["Moon", "Earth", "Jupiter", "Neptune", "Earth", "Venus"]
get_indexes = lambda x, xs: [i for (y, i) in zip(xs, range(len(xs))) if x == y]
print(get_indexes("Earth", x))

以下代码提取满足条件的峰值：

intensitySortedPeaks = sorted(peaks, key=lambda p: p.Intensity, reverse=True)
for i in range(len(intensitySortedPeaks)):
    testMass = intensitySortedPeaks[i].MZ + 1.0033
    results = [p for p in intensitySortedPeaks if abs(GetPPMError(p.MZ,testMass)) < 10]
 ###how do we get the indexes saved as indeResults

如何获取存储在结果变量中的元素的索引。我想从循环

中的原始intensitySortedList中删除这些元素

如何修改上述代码才能完成此任务？

Answer 1

如果只想要匹配项目的索引，请尝试：

def indices_satisfying(fn, lst):
  return [index for index,x in enumerate(lst) if fn(x)]

如果你想删除令人满意的元素吗

def removing_satisfying(fn, lst]:
  return [x for x in lst if not fn(x)]

获取搜索结果的索引

1 个答案: