Given 2 list of integers how to find the non-overl

问题:

Given

x = [5, 30, 58, 72]
y = [8, 35, 53, 60, 66, 67, 68, 73]

The goal is to iterate through x_i and find the value for y that's larger than x_i but not larger than x_i+1

Assume that both list are sorted and all items are unique, the desired output given the x and y is:

[(5, 8), (30, 35), (58, 60), (72, 73)]

I've tried:

def per_window(sequence, n=1):
    """
    From http://stackoverflow.com/q/42220614/610569
        >>> list(per_window([1,2,3,4], n=2))
        [(1, 2), (2, 3), (3, 4)]
        >>> list(per_window([1,2,3,4], n=3))
        [(1, 2, 3), (2, 3, 4)]
    """
    start, stop = 0, n
    seq = list(sequence)
    while stop <= len(seq):
        yield tuple(seq[start:stop])
        start += 1
        stop += 1

x = [5, 30, 58, 72]
y = [8, 35, 53, 60, 66, 67, 68, 73]

r = []

for xi, xiplus1 in per_window(x, 2):
    for j, yj in enumerate(y):
        if yj > xi and yj < xiplus1:
            r.append((xi, yj))
            break

# For the last x value.
# For the last x value.
for j, yj in enumerate(y):
    if yj > xiplus1:
        r.append((xiplus1, yj))
        break

But is there a simpler way to achieve the same with numpy, pandas or something else?

回答1:

You can use numpy.searchsorted with side='right' to find out the index of the first value in y that is larger than x and then extract the elements with the index; A simple version which assumes there is always one value in y larger than any element in x could be:

x = np.array([5, 30, 58, 72])
y = np.array([8, 35, 53, 60, 66, 67, 68, 73])

np.column_stack((x, y[np.searchsorted(y, x, side='right')]))
#array([[ 5,  8],
#       [30, 35],
#       [58, 60],
#       [72, 73]])

---

Given y is sorted:

np.searchsorted(y, x, side='right')
# array([0, 1, 3, 7])

returns the index of the first value in y that is larger than the corresponding value in x.

回答2:

We can use pd.DataFrame on list with merge_asof with direction = forward i.e

new = pd.merge_asof(pd.DataFrame(x,index=x), pd.DataFrame(y,index=y),on=0,left_index=True,direction='forward')
out = list(zip(new[0],new.index))

If you dont need exact matches to match the you need to pass allow_exact_matches=False to merge_asof

Output :

[(5, 8), (30, 35), (58, 60), (72, 73)]

回答3:

You can construct a new list by iterating over x zipped with itself -- offset by 1 index and appended with the last element of y -- and then iterating over y, check the condition at each pass and break the inner most loop.

out = []
for x_low, x_high in zip(x, x[1:]+y[-1:]):
    for yy in y:
        if (yy>x_low) and (yy<=x_high):
            out.append((x_low,yy))
            break

out
# returns:
[(5, 8), (30, 35), (58, 60), (72, 73)]

回答4:

def find(list1,list2):
    final = []
    for i in range(len(list1)):
        pos=0
        try:
            while True:
                if i+1==len(list1) and list1[i]<list2[pos]:
                    final.append((list1[i],list2[pos]))
                    raise Exception
                if list1[i]<list2[pos] and list1[i+1]>list2[pos]:
                    final.append((list1[i],list2[pos]))
                    raise Exception
                pos+=1
        except: pass
    return final