查找lis中每个嵌套列表的最小值和最大值

Question

我正在尝试查找嵌套列表中每个列表的最小值和最大值，以及最小值或最大值出现的时间的索引：

例如：

l=[[5,6,7][6,10,9,6][2,3,1]]

变成：

maxl=[7,10,3]
indexl=[2,1,2]

我试过这个，它似乎让我最大的名单（没有索引尚未），但不为最小-有人知道如何最好地做到这一点？你知道吗

maxHap=[]
for subL in happiness1:
    maxHap.append(max(subL))
print(maxHap)

minHap=[]
for subL in happiness1:
    minHap.append(min(subL))
print(minHap)

一个新手谢谢你

Answer 1

可以使用方法index获取列表值的索引： https://docs.python.org/3/tutorial/datastructures.html

maxl = []
indexl = []
for inner_list in l:
    max_value = max(inner_list)
    maxl.append(max(max_value))
    indexl.append(inner_list.index(max_value)

Answer 2

其他选项：

l=[[5,6,7],[6,10,9,6],[2,3,1]]

maxs = [ max(s) for s in l ] #=> [7, 10, 3]
max_idxs = [ s.index(max(s)) for s in l ] #=> [2, 1, 1]
mins = [ min(s) for s in l ] #=> [5, 6, 1]
mins_idxs = [ s.index(min(s)) for s in l ] #=> [0, 0, 2]

---

也可以将结果存储到dict数组中，一行：

mapp = map(lambda x: {'max': max(x), 'max_idxs': x.index(max(x)), 'min': min(x), 'min_idxs': x.index(min(x)) }, l)

for k in mapp:
  print(k)

#=> {'max': 7, 'max_idxs': 2, 'min': 5, 'min_idxs': 0}
#=> {'max': 10, 'max_idxs': 1, 'min': 6, 'min_idxs': 0}
#=> {'max': 3, 'max_idxs': 1, 'min': 1, 'min_idxs': 2}

Answer 3

您可以使用以下一种内衬：

l=[[5,6,7],[6,10,9,6],[2,3,1]]
maxL, index = zip(*[(max(subList), subList.index(max(subList))) for subList in l])
print(maxL) # will output (7, 10, 3)
print(index) # will output (2, 1, 1)