为 Python 中包含 [] 个元素的多个列表的元素定义总和

Question

_你好，世界！我在从多个列表中删除元素时遇到问题。我有多个列表。

print("Preparation")
print(dateTableMonthValues)

它打印我：

[['4:10'], ['3:20'], ['3:45'], ['32:05'], ['5:00'], ['5:00'], ['5:00'], ['5:00'], ['5:00'], ['5:00'], ['5:00'], ['35:00'], ['3:45'], ['5:00'], ['5:00'], ['2:30'], ['2:30'], ['3:45'], ['6:15'], ['28:45'], ['5:00'], ['5:00'], ['7:55'], ['7:05'], ['5:00'], ['5:50'], [], ['35:50']]

我需要从这个数组中删除几个元素，例如 ['32:05'], ['35:00'], ['28:45'], [], ['35:50' ]]

好的，我找到了这些值并且：

for i in range(len(dateTableDayValues)):
    week = dateTableDayValues[i][0]
    print(dateTableDayValues[i][0])
    if week[3] == 'w':
        print('w')
        dateTableMonthValues[i].pop(0)
print("Check")
print(dateTableMonthValues)

现在它打印我：

[['4:10'], ['3:20'], ['3:45'], [], ['5:00'], ['5:00'], ['5:00'], ['5:00'], ['5:00'], ['5:00'], ['5:00'], [], ['3:45'], ['5:00'], ['5:00'], ['2:30'], ['2:30'], ['3:45'], ['6:15'], [], ['5:00'], ['5:00'], ['7:55'], ['7:05'], ['5:00'], ['5:50'], [], []]

所以，我有两个问题。首先，为什么方法 'pop' 没有删除元素，因为我仍然有这样的元素：'[]'。其次，如何从数组（列表）中删除这些用于定义所有元素总和的空元素“[]”？它不起作用：

if dateTableMonthValues[i][0] == []:
         del dateTableMonthValues[i][0]
         dateTableMonthValues[i].pop(0)

老问题：因此，当我尝试访问 list[3][0] 时，它给我写了“list index out of range”。所以我决定删除该元素。

但无论如何我无法定义所有元素的总和。我仍然继续收到消息

IndexError: list index out of range

请帮帮我，我该如何解决这个问题？

Answer 1

有很多方法可以解决！

编辑：（问题稍作修改）

为什么方法 'pop' 没有删除元素？因为你访问了元素'i'，删除了他的第一个元素。如果你想删除它，你可以这样做：

dateTableMonthValues.pop(i) # Not: dateTableMonthValues[i].pop(0)

或

dateTableMonthValues.remove(the_element)

如果你想删除列表中的每个 [ ]，你可以这样做：

lst = [['4:10'], ['3:20'], ['3:45'], [], ['5:00'], ['5:00'], ['5:00'], ['5:00'], ['5:00'], ['5:00'], ['5:00'], [], ['3:45'], ['5:00'], ['5:00'], ['2:30'], ['2:30'], ['3:45'], ['6:15'], [], ['5:00'], ['5:00'], ['7:55'], ['7:05'], ['5:00'], ['5:50'], [], [],[],[],[]]
i = 0
while i<len(lst):
    if lst[i] == []:
        lst.pop(i)
    else: 
        i+=1
print(lst) #[['4:10'], ['3:20'], ['3:45'], ['5:00'], ['5:00'], ['5:00'], ['5:00'], ['5:00'], ['5:00'], ['5:00'], ['3:45'], ['5:00'], ['5:00'], ['2:30'], ['2:30'], ['3:45'], ['6:15'], ['5:00'], ['5:00'], ['7:55'], ['7:05'], ['5:00'], ['5:50']]

枚举

l = [['4:10'], ['3:20'], ['3:45'], ['Delete me'], ['5:00'], ['Delete me'], ['10:00']]

for i,e in enumerate(l):
    if e == ["Delete me"]:
        l.pop(i)

print(l) #[['4:10'], ['3:20'], ['3:45'], ['5:00'], ['10:00']]

但是如果你真的不知道如何使用enumerate:

Try/Except

l = [['4:10'], ['3:20'], ['3:45'], ['Delete me'], ['5:00'], ['Delete me'], ['10:00']]

for i in range(len(l)):
    try:
        if l[i] == ["Delete me"]:
            l.pop(i)
    except IndexError:
        pass

print(l) #[['4:10'], ['3:20'], ['3:45'], ['5:00'], ['10:00']]

为了简单起见：当出现“IndexError”时，程序只是 pass，因为什么都没发生。（实际上，从系统的角度来看它更复杂，但你可以这样解释！）

移除

l = [['4:10'], ['3:20'], ['3:45'], ['Delete me'], ['5:00'], ['Delete me'], ['10:00']]

for e in l:
    if e == ["Delete me"]:
        l.remove(e)

print(l) #[['4:10'], ['3:20'], ['3:45'], ['5:00'], ['10:00']]

最后但并非最不重要的一点！ list.remove(element) 从列表中删除元素，而不是像 pop

这样的索引的元素

为 Python 中包含 [] 个元素的多个列表的元素定义总和

Define sum for elements of multiple list with [] elements in Python

python

list

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