代码之家 › 专栏 › 技术社区 › Jane Sully

python中数组的就地修改

in-place indexing arrays python

0

Jane Sully · 技术社区 · 6 年前

我发现这个问题要求对数组进行就地修改,以便将所有零移动到数组的末尾,并保持非零元素的剩余顺序。根据问题陈述,就地意味着不复制原始数组(这是从Leetcode中提取的,可以找到#283,移动零)

for num in nums:
    if num == 0:
        nums.remove(num)
        nums.append(0)

解决方案清晰易懂,所以我知道它在做它应该做的事情。

indices = []
for en, num in enumerate(nums): # get the index of all elements that are 0
    if num == 0:
        indices.append(en)

for en, i in enumerate(indices): 
    new_i = i-en # use the index, accounting for the change in length of the array from removing zeros
    nums = nums[:new_i] + nums[new_i+1:] # remove the zero element
nums = nums + [0] * len(indices) # add back the appropriate number of zeros at the end

1 回复 | 直到 6 年前

1

0

bunbun 6 年前

remove是否在内部复制数组以删除指定的元素?

source code for lists , remove() 电话 listremove() :

listremove(PyListObject *self, PyObject *v)
{
    Py_ssize_t i;

    for (i = 0; i < Py_SIZE(self); i++) {
        int cmp = PyObject_RichCompareBool(self->ob_item[i], v, Py_EQ);
        if (cmp > 0) {
            if (list_ass_slice(self, i, i+1, (PyObject *)NULL) == 0)
                Py_RETURN_NONE;
            return NULL;
        }
        else if (cmp < 0)
            return NULL;
    }
    PyErr_SetString(PyExc_ValueError, "list.remove(x): x not in list");
    return NULL;
}

here :

arguments: list object, element to remove
returns none if OK, null if not
listremove:
    loop through each list element:
        if correct element:
            slice list between element's slot and element's slot + 1
            return none
    return null