Python-填補列表中的空白

Question

我有一個由元組組成的列表，是從數據庫查詢生成的，例如：

list = [(0,1,1), (1,2,1), (2,4,3), (4,2,1)]

每個元組中的第一個數字必須是一個連續的數字，從0到15。可能還會有一些缺失的數字，我正在尋找填補空白的最佳方法。

目前，我是通過循環來完成此操作的，但是作為Python noob，我認為它很草率，並且有更好的方法：

# first fill in gaps
cnt = 0
for a,b,c in list:
    if a > cnt:
        list.insert(cnt, tuple((cnt, 0, 0)))
    cnt += 1

# then add any missing at end
while cnt < 16:
    list.append(tuple((cnt, 0, 0)))
    cnt += 1

因此，列表開始時的預期輸出為：

list = [(0,1,1), (1,2,1), (2,4,3), (3,0,0), (4,2,1), (5,0,0), (6,0,0), (7,0,0), (8,0,0), (9,0,0), (10,0,0), (11,0,0), (12,0,0), (13,0,0), (14,0,0), (15,0,0)]

Answer 1

有很多方法，您可以像這樣生成一個新列表：

data = [(0,1,1), (1,2,1), (2,4,3), (4,2,1)]

out = []
for i in range(16):
    if data and i == data[0][0]:
        out.append(data.pop(0))
    else:
        out.append((i, 0, 0))

print(out)
# [(0, 1, 1), (1, 2, 1), (2, 4, 3), (3, 0, 0), (4, 2, 1), 
# (5, 0, 0), (6, 0, 0), (7, 0, 0), (8, 0, 0), (9, 0, 0), 
# (10, 0, 0), (11, 0, 0), (12, 0, 0), (13, 0, 0), (14, 0, 0), (15, 0, 0)]

附帶說明一下，我重命名了列表data ，因為最好避免將內置函數的名稱用作變量。

Answer 2

您可以將元組列表轉換為首先由元組的第一項索引的字典，以便您可以迭代0到15的范圍以找到缺失的索引，並在列表理解中為其生成默認值：

l = [(0,1,1), (1,2,1), (2,4,3), (4,2,1)]
d = {k: v for k, *v in l}
print([(i, *d.get(i, (0, 0))) for i in range(16)])

輸出：

[(0, 1, 1), (1, 2, 1), (2, 4, 3), (3, 0, 0), (4, 2, 1), (5, 0, 0), (6, 0, 0), (7, 0, 0), (8, 0, 0), (9, 0, 0), (10, 0, 0), (11, 0, 0), (12, 0, 0), (13, 0, 0), (14, 0, 0), (15, 0, 0)]

Answer 3

為了介紹一些有趣的語言功能（可用於解決此問題），這里提供了一種基於Python穩定排序的解決方案，該解決方案具有簡單的排序鍵函數和itertools.groupby() ，可根據索引對項目進行分組。

然后，如果輸入數據中有此索引，則取兩個中的第一個。 如果沒有此索引，則為默認項。

import itertools

received_items = [(0,1,1), (1,2,1), (2,4,3), (4,2,1)]
print(received_items)

default_items = [(i, 0, 0) for i in range(16)]

# append default items at the end
data = received_items + default_items
# perform stable sort on the index (first element) selected via key function
keyfunc = lambda x: x[0]
data.sort(key=keyfunc)
# always take first item for each index group
out = [next(v) for k,v in itertools.groupby(data, key=keyfunc)]

print(out)

輸出為：

[(0, 1, 1), (1, 2, 1), (2, 4, 3), (4, 2, 1)]
[(0, 1, 1), (1, 2, 1), (2, 4, 3), (3, 0, 0), (4, 2, 1), (5, 0, 0), (6, 0, 0), (7, 0, 0), (8, 0, 0), (9, 0, 0), (10, 0, 0), (11, 0, 0), (12, 0, 0), (13, 0, 0), (14, 0, 0), (15, 0, 0)]

以供參考：

• https://docs.python.org/3/library/stdtypes.html#list.sort
• https://docs.python.org/3/library/itertools.html#itertools.groupby
• https://docs.python.org/3/reference/expressions.html#lambda