Question

我有以下列表

['0, 0, 0.16548735788092,\n',
 '1, 3.90625E-05, 0.368149412097409,\n',
 '2, 7.8125E-05, 0.184674297925085,\n',
 '3, 0.0001171875, 0.00359755125828087,\n',
 '4, 0.00015625, 0.0131910212803632,\n',
 '5, 0.0001953125, 0.24703185306862,\n',
 '6, 0.000234375, 0.474876766093075,\n',]

BWhich我是从

获得的

data = f.readlines()

我想将此列表转换为包含3列的数组。感谢

Answer 1

我们按行重复。然后，对于每一行，我们在,上拆分并将相应的值分配给临时值，然后将一个包含三个条目的列添加到数组

array=[]
for line in f.readlines():
    a_0,a_1,a_2,_=line.split(",")
    array.append([a_0,a_1,a_2])

Answer 2

[list(map(float, i.strip().split(',')[:-1])) for i in mylist]

输出：

[[0.0, 0.0, 0.16548735788092],
 [1.0, 3.90625e-05, 0.368149412097409],
 [2.0, 7.8125e-05, 0.184674297925085],
 [3.0, 0.0001171875, 0.00359755125828087],
 [4.0, 0.00015625, 0.0131910212803632],
 [5.0, 0.0001953125, 0.24703185306862],
 [6.0, 0.000234375, 0.474876766093075]]

Answer 3

如果您希望将列表作为数字，请转到：

list_3d = []

for line in data:
    columns = list(map(float, line.replace(' ','').split(',')[:-1]))
    list_3d.append(columns)

print(list_3d)

Answer 4

您可以尝试这样的事情：

data=['0, 0, 0.16548735788092,\n',
 '1, 3.90625E-05, 0.368149412097409,\n',
 '2, 7.8125E-05, 0.184674297925085,\n',
 '3, 0.0001171875, 0.00359755125828087,\n',
 '4, 0.00015625, 0.0131910212803632,\n',
 '5, 0.0001953125, 0.24703185306862,\n',
 '6, 0.000234375, 0.474876766093075,\n',]



print(list(map(lambda x:list(map(float,x.split(',')[:-1])),data)))

输出：

[[0.0, 0.0, 0.16548735788092], [1.0, 3.90625e-05, 0.368149412097409], [2.0, 7.8125e-05, 0.184674297925085], [3.0, 0.0001171875, 0.00359755125828087], [4.0, 0.00015625, 0.0131910212803632], [5.0, 0.0001953125, 0.24703185306862], [6.0, 0.000234375, 0.474876766093075]]

Answer 5

您可以使用string.split，', '作为分隔符。

new_list = []

for line in data:
    line = line.strip()      # get rid of \n
    line = line.strip(',')   # get rid of trailing comma
    line = line.split(', ')  # split into items
    new_list.append(line)    # add to the new list

然后输出：

[['0', '0', '0.16548735788092'],
 ['1', '3.90625E-05', '0.368149412097409'],
 ...

Answer 6

这是一种方式。输入列表为lst。输出是3个元组的列表（每列一个）。我还将数字转换为float并删除了\n。

cols = list(zip(*(map(float, i.replace('\n', '').split(',')[:-1]) for i in lst)))

Answer 7

正如我从这个问题中理解的那样，需要2D（mx3）数组，它存储数据变量的每个元素。

#creating 2-D Matrix 
Matrix = [[0 for x in range(0,3)] for y in range(0,len(data))]
    for x in data :
        a1=x.split(",") #split each element
        for i in range(0,len(data)):
            for j in range(0,3): #removes inserting \n
                  Matrix[i][j]=a1[j] 
print(Matrix)

将多个元素列表转换为数组

7 个答案: