我试图遍历列表列表。我想在列表中找到列的最小值和最大值。
这是列表:
['Afghanistan', 2.66171813, 7.460143566, 0.490880072, 52.33952713, 0.427010864, -0.106340349, 0.261178523]
['Albania', 4.639548302, 9.373718262, 0.637698293, 69.05165863, 0.74961102, -0.035140377, 0.457737535]
['Algeria', 5.248912334, 9.540244102, 0.806753874, 65.69918823, 0.436670482, -0.194670126, 0]
['Argentina', 6.039330006, 9.843519211, 0.906699121, 67.53870392, 0.831966162, -0.186299905, 0.305430293]
['Armenia', 4.287736416, 9.034710884, 0.697924912, 65.12568665, 0.613697052, -0.132166177, 0.246900991]
['Australia', 7.25703764, 10.71182728, 0.949957848, 72.78334045, 0.910550177, 0.301693261, 0.45340696]
我试图遍历从li[1]
到li[-1]
的每一列。
我尝试过:
j = 1
small = []
while j < len(li)
for i in range(0, len(newlist)):
small.append(newlist[i][j])
print(max(small))
print(min(small))
j = j + 1
输出:
7.25703764
2.66171813
我得到第一列,但我正努力遍历下几列。
答案 0 :(得分:1)
您可以使用zip
。给您的数据集为my_list
:
my_lists = [['Afghanistan', 2.66171813, 7.460143566, 0.490880072, 52.33952713, 0.427010864, -0.106340349, 0.261178523],
['Albania', 4.639548302, 9.373718262, 0.637698293, 69.05165863, 0.74961102, -0.035140377, 0.457737535],
['Algeria', 5.248912334, 9.540244102, 0.806753874, 65.69918823, 0.436670482, -0.194670126, 0],
['Argentina', 6.039330006, 9.843519211, 0.906699121, 67.53870392, 0.831966162, -0.186299905, 0.305430293],
['Armenia', 4.287736416, 9.034710884, 0.697924912, 65.12568665, 0.613697052, -0.132166177, 0.246900991],
['Australia', 7.25703764, 10.71182728, 0.949957848, 72.78334045, 0.910550177, 0.301693261, 0.45340696]]
for t in list(zip(*my_lists))[1:]:
print(max(t), min(t))
7.25703764 2.66171813
10.71182728 7.460143566
0.949957848 0.490880072
72.78334045 52.33952713
0.910550177 0.427010864
0.301693261 -0.194670126
0.457737535 0
答案 1 :(得分:0)
columns=len(newlist[0])
rows=len(newlist)
for i in range(1,columns):
for j in range(rows):
small.append(newlist[j][i])
print("{}. columuns min value:{} max value:{}".format(i,min(small),max(small)))
答案 2 :(得分:0)
下面的代码使用列表理解功能创建一个具有每个列表的最小值和最大值的元组列表,您可以进一步格式化以在每个元组中也具有国家/地区名称。
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