Python3 数据插补

Question

我轰炸的求职面试。

    Remove all rows where at least half of the entries are negative
    Fill in remaining negative values in each column with the 
    average for that column, excluding invalid entries

Input: [[5],
     [3],
     [1.0, 2.0, 10.0],
     [-1.0, -99.0, 0],
     [-1.0, 4.0, 0],
     [3.0, -6.0, -0.1],
     [1.0, -0.31, 6.0]
    ]

Output: new mean rounded to one decimal place

不知道从哪里开始

输出新平均值，四舍五入到小数点后一位

Answer 1

假设有 3 列，并且在剩余的每一列中有 1 个或没有负值被整个列的平均值替换。

然后：

np = [[5],[3],[1.0,2.0,10.0],[-1.0,-99.0,0],[-1.0,4.0,0],[3.0,-6.0,-0.1],[1.0,-0.31, 6.0]]
cols = 0  # save num of cols for later
for l in inp:
    pos = 0  # count positive
    neg = 0  # count negative
    for n in l:
        if n > 0:
            pos += 1  # update
        elif n < 0:
            neg += 1  # update
    if pos+neg > cols:  # save num of cols
        cols = pos+neg
    if pos < neg:  # remove list with too many negatives
        inp.remove(l)

for i in range(cols):  # loop through cols
    neg_index = 0  # find the negative value's index to replace with the average
    entries = 0  # for calculating the average
    summ = 0  # for calculating the average
    for c in inp:
        try:  # see if col exist in a list
            if c[i] < 0:
                neg_index = inp.index(c)  # save index of negative value found
            else:
                summ += c[i]
                entries += 1
        except:
            continue
    try:  # see if col exist in list
        inp[neg_index][i] = round(summ / entries,1)  # replace negative value index with average
    except:
        continue

print(inp)

结果：

[5],
[3],
[1.0, 2.0, 10.0],
[2.5, 4.0, 0],
[1.0, 3.0, 6.0],

我相信这就是他们正在寻找的东西，希望这会有所帮助。