從列表中刪除匹配項

Question

我有一個值列表。 我想刪除相互抵消的值（+ 和 -）。 這些值在列表中是隨機的，所以我首先在 excel 中添加了一個包含絕對值的新列。 然后我對絕對值進行排序，因此需要取消的金額低於彼此。

我想創建一個 for 循環並將第一行與第二行相加，當總和為零時，刪除兩行並再次從頂部開始。 請參考示例圖片。 我已將應刪除的項目標記為黃色。 由於我只想刪除匹配項，操作后amount列的總和應該不會改變。

目前我有這樣的東西

for i in df["實體貨幣金額"]: if df["實體貨幣金額"][i] + df["實體貨幣金額"][i+1] == 0: df.drop(df [df["實體貨幣金額"][i]]) df.drop(df[df["實體貨幣金額"][i + 1]])

Answer 1

在對列表進行排序后嘗試這樣的操作（正如您已經說過的）：

for i,elem in enumerate(yourList):
    nextElem = yourList[i+1]
    if (elem + nextElem < 0.00000001): 
        yourList.remove(elem)
        yourList.remove(nextElem)

Answer 2

我會根據幾個構建塊來回答。 首先是創建兩個“可迭代”子列表。 一個用於正數，另一個用於負數。

然后我將使用next()遍歷它們兩個，只要兩個列表之一有值，我就會適當地對當前值采取行動。

import random

full_data = [random.randint(0, 10) for _ in range(20)] + [-random.randint(0, 10) for _ in range(20)]

zeros = [i for i in full_data if i == 0]
positives = iter(sorted([i for i in full_data if i > 0]))
negatives = iter(sorted([i for i in full_data if i < 0], reverse=True))

## ------------------------------
## prime the list with zero or 1 0s where an odd number of 0s
## results in [0] as the evens cancel each other out.
## ------------------------------
result = [0] * (len(zeros) % 2)
## ------------------------------

pos = next(positives, None)
neg = next(negatives, None)
while pos is not None or neg is not None:
    if pos is None:
        # we ran out of positive numbers so add any remaining negatives
        result.extend([neg] + list(negatives))
        break

    if neg is None:
        # we ran out of negative numbers so add any remaining positives
        result.extend([pos] + list(positives))
        break

    if pos == -neg:
        # these results cancel each other
        pos = next(positives, None)
        neg = next(negatives, None)

    elif pos > -neg:
        # this positive is "larger" then this negative so add the negative
        result.append(neg)
        neg = next(negatives, None)

    else:
        # this positive is "smaller" than this negative so add the positive
        result.append(pos)
        pos = next(positives, None)

print(f"The original list has {len(full_data)} items and sums to: {sum(full_data)}")
print(f"The filtered list has {len(result)} items and sums to: {sum(result)}")