Question

我在python中有一些整数列表：

[[2, 8, 10, 500, 502], [1, 4, 5, 401]]

如何根据列表中数字之间的差异将值扩展为连续范围，以便得到如下内容：

[[2, 3, 4, 5, 6, 7, 8, 9, 10, 500, 501, 502], [1, 2, 3, 4, 5, 401]]

所以，基本上，如果列表中的项目之间的差异小于100，则只将一组数字扩展为一个完整范围？

Answer 1

这很难看，但试试这个：

def list_expand(x):
    new_list = []
    while True:
        if len(x) < 2:
            new_list.append(x[0])
            break

        m = min(x)
        x.remove(m)
        if abs(m - min(x)) < 100:
            new_list.extend(range(m, min(x)))
        else:
            new_list.append(m)
    return new_list

它通过了这些测试：

assert list_expand([99, 0, 198]) == range(0, 199)
assert list_expand([100, 0, 200]) == [0, 100, 200]
assert list_expand([2, 8, 10, 500, 502]) == range(2, 11) + range(500, 503)
assert list_expand([1, 4, 5, 401]) == range(1, 6) + [401]

Answer 2

这是另一个版本，可能更短：

In [7]: from collections import OrderedDict

In [107]: def foo(l):
     ...:     l=sorted(l)
     ...:     t = l + [e for i, v in enumerate(l[1:]) for e in range(l[i], v) if v-l[i]<100]
     ...:     return list(OrderedDict.fromkeys(sorted(t)))

In [108]: assert foo([99, 0, 198]) == range(0, 199)
     ...: assert foo([100, 0, 200]) == [0, 100, 200]
     ...: assert foo([2, 8, 10, 500, 502]) == range(2, 11) + range(500, 503)
     ...: assert foo([1, 4, 5, 401]) == range(1, 6) + [401]

Answer 3

用itertools.izip_longest和三元一样的操作被黑客攻击：）

from itertools import izip_longest as zip

def list_expand(c):
   c = sorted(c)
   return [k for i,j in zip(c,c[1:],fillvalue=0) for k in [[i],range(i,j)][0<=j-i<100]]

assert list_expand([99, 0, 198]) == range(0, 199)
assert list_expand([100, 0, 200]) == [0, 100, 200]
assert list_expand([2, 8, 10, 500, 502]) == range(2, 11) + range(500, 503)
assert list_expand([1, 4, 5, 401]) == range(1, 6) + [401]

Answer 4

a = [2,8,10,500,502]

def expand_list(a):
    for element in a:
        if element < 10:
            for i in range(element):
                if i not in a:
                    a.append(i)
    return sorted(a)

print expand_list(a)

这可能不是最佳解决方案，但我把这作为时间挑战;）

编辑：这应该通过测试，虽然我不确定它是否正确。但我想我会留下它。

a = [8, 2, 10, 500, 502]

def expand_list(a):
    a = sorted(a)
    for i in range(len(a)):
        if i > 0:
            if (a[i] - a[i-1]) < 100:
                for j in range(a[i-1], a[i]):
                    if j not in a:
                        a.append(j)
    return sorted(a)
print expand_list(a)

Answer 5

这应该有效：

v = [100, 0, 200]
v = sorted(v)
exp = []
for i, ele in enumerate(v[1:]):
    if ele - v[i] < 100:
        exp.extend(range(v[i], ele))
        exp.extend([ele]) # to add the last element of the range
if not exp:
    exp = v  # to take care of cases where the numbers in original list are all separated by > 100
else:
    exp = list(set(exp)) # to get rid of the duplicates had the input been [2, 8, 10, 11, 500, 502]
print exp

如何根据阈值扩展整数列表？

5 个答案: