根據元素的某些方面，如何將Python列表分成兩個列表

Question

我有一個這樣的列表：

[[8, "Plot", "Sunday"], [1, "unPlot", "Monday"], [12, "Plot", "Monday"], [10, "Plot", "Tuesday"], [4, "unPlot", "Tuesday"], [14, "Plot", "Wednesday"], [6, "unPlot", "Wednesday"], [1, "unPlot", "Thursday"], [19, "Plot", "Thursday"], [28, "Plot", "Friday"], [10, "unPlot", "Friday"], [3, "unPlot", "Saturday"]]

我想根據Plot和unPlot值將它分成兩個列表，結果如下：

list1=[[8, "Plot", "Sunday"], [12, "Plot", "Monday"], ...]
list2=[[1, "unPlot", "Monday"], [4, "unPlot", "Tuesday"], ...]

Answer 1

嘗試基本列表理解：

>>> [ x for x in l if x[1] == "Plot" ]
[[8, 'Plot', 'Sunday'], [12, 'Plot', 'Monday'], [10, 'Plot', 'Tuesday'], [14, 'Plot', 'Wednesday'], [19, 'Plot', 'Thursday'], [28, 'Plot', 'Friday']]
>>> [ x for x in l if x[1] == "unPlot" ]
[[1, 'unPlot', 'Monday'], [4, 'unPlot', 'Tuesday'], [6, 'unPlot', 'Wednesday'], [1, 'unPlot', 'Thursday'], [10, 'unPlot', 'Friday'], [3, 'unPlot', 'Saturday']]

如果您喜歡函數式編程，請使用filter ：

>>> filter(lambda x: x[1] == "Plot", l)
[[8, 'Plot', 'Sunday'], [12, 'Plot', 'Monday'], [10, 'Plot', 'Tuesday'], [14, 'Plot', 'Wednesday'], [19, 'Plot', 'Thursday'], [28, 'Plot', 'Friday']]
>>> filter(lambda x: x[1] == "unPlot", l)
[[1, 'unPlot', 'Monday'], [4, 'unPlot', 'Tuesday'], [6, 'unPlot', 'Wednesday'], [1, 'unPlot', 'Thursday'], [10, 'unPlot', 'Friday'], [3, 'unPlot', 'Saturday']]

我個人覺得列表理解更加清晰。 它肯定是最“pythonic”的方式。

Answer 2

data = [[8, "Plot", "Sunday"], [1, "unPlot", "Monday"], [12, "Plot", "Monday"], [10, "Plot", "Tuesday"], [4, "unPlot", "Tuesday"], [14, "Plot", "Wednesday"], [6, "unPlot", "Wednesday"], [1, "unPlot", "Thursday"], [19, "Plot", "Thursday"], [28, "Plot", "Friday"], [10, "unPlot", "Friday"], [3, "unPlot", "Saturday"]]

res = {'Plot':[],'unPlot':[]}
for i in data: res[i[1]].append(i)

這樣您可以迭代列表一次

Answer 3

嘗試：

yourList=[[8, "Plot", "Sunday"], [1, "unPlot", "Monday"], [12, "Plot", "Monday"], [10, "Plot", "Tuesday"], [4, "unPlot", "Tuesday"], [14, "Plot", "Wednesday"], [6, "unPlot", "Wednesday"], [1, "unPlot", "Thursday"], [19, "Plot", "Thursday"], [28, "Plot", "Friday"], [10, "unPlot", "Friday"], [3, "unPlot", "Saturday"]]
plotList=[]
unPlotList=[]

for i in yourList:
    if "Plot" in i:
        plotList.append(i)
    else:
        unPlotList.append(i)

理解時更短或更短：

plotList = [i for i in yourList if "Plot" in i]
unPlotList = [i for i in yourList if "unPlot" in i]

Answer 4

您可以使用列表推導，例如

# old_list elements should be tuples if they're fixed-size, BTW
list1 = [(X, Y, Z) for X, Y, Z in old_list if Y == 'Plot']
list2 = [(X, Y, Z) for X, Y, Z in old_list if Y == 'unPlot']

如果只想遍歷輸入列表一次，那么可能：

def split_list(old_list):
    list1 = []
    list2 = []
    for X, Y, Z in old_list:
        if Y == 'Plot':
            list1.append((X, Y, Z))
        else:
            list2.append((X, Y, Z))
    return list1, list2

Answer 5

您可以直接瀏覽列表，並檢查值是否為“Plot”，如下所示：

for i in List:
  if i[1]=="Plot":
    list1.append(i)
  else:
    list2.append(i)

Answer 6

我有一個輔助函數，用於將列表分成兩部分的一般情況：

def partition(iterable, condition):
        def partition_element(partitions, element):
            (partitions[0] if condition(element) else partitions[1]).append(element)
            return partitions
        return reduce(partition_element, iterable, ([], []))

例如：

>>> partition([1, 2, 3, 4], lambda d: d % 2 == 0)
([2, 4], [1, 3])

或者在你的情況下：

>>> partition(your_list, lambda i: i[1] == "Plot")

Answer 7

使用列表理解：

l = [[8, "Plot", "Sunday"], [1, "unPlot", "Monday"], [12, "Plot", "Monday"], [10, "Plot", "Tuesday"], [4, "unPlot", "Tuesday"], [14, "Plot", "Wednesday"], [6, "unPlot", "Wednesday"], [1, "unPlot", "Thursday"], [19, "Plot", "Thursday"], [28, "Plot", "Friday"], [10, "unPlot", "Friday"], [3, "unPlot", "Saturday"]]

list1 = [x for x in l if x[1] == "Plot"]

list2 = [x for x in l if x[1] == "unPlot"]

Answer 8

您也可以使用filter命令執行此操作：

list1 = filter(lambda x: x[1] == "Plot", list)
list2 = filter(lambda x: x[1] == "unPlot", list)