如何Python列表分爲兩個列表，按照元素的某些方面

我有一個這樣的名單：如何Python列表分爲兩個列表，按照元素的某些方面

[[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"]]

我想把它分成根據Plot和unPlot值兩份名單，導致：

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

來源

2012-01-30 saun jean

嘗試使用基本列表理解：

>>> [ 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']]

或用filter，如果你看中了函數式編程：

>>> 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perso最終找到更清晰的列表理解。這當然是最「pythonic」的方式。

來源

2012-01-30 14:30:20 Chewie

感謝這樣一個簡單的解決方案。 – 2012-01-30 14:42:37

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)

或更短與補償rehension：

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

來源

2012-01-30 14:29:44 ezdazuzena

ezdazuzena明白了......應該工作......只要確保先將plotlist和unplotlist定義爲空列表即可。 – maneesha 2012-01-30 14:32:44

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"]

來源

2012-01-30 14:29:56 soulcheck

您可以使用列表解析，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']

如果你想遍歷輸入列表中只有一次，那麼也許：

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

來源

2012-01-30 14:30:32

你可以簡單地瀏覽清單，並檢查是否值是「暗算」是這樣的：

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

來源

2012-01-30 14:30:58 MartinHaTh

你也可以用filter命令做到這一點：

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

來源

2012-01-30 14:32:07

我有分區列表一般情況下的輔助函數有兩種：

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])

或者你的情況：

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

來源

2012-01-30 14:33:42 wutz

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)

這樣你迭代一次列表

來源

2012-01-30 14:34:06

如何Python列表分爲兩個列表，按照元素的某些方面

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