如何从 python 字典中替换 2D 列表值？答案

【问题标题】：How to replace 2D list values from python dictionary?如何从 python 字典中替换 2D 列表值？
【发布时间】：2019-02-23 03:32:04
【问题描述】：

我正在运行时创建一个 Python 字典，如下所示，

valueDict =  {(0, 0): 'P0', (20, 0): 'P1', (20, 11.36): 'P2', (0, 11.36): 'P3'}

我有两个数组；

values = [[0, 0, 20, 0, 15, 5.5, 5, 5.5]
    [20, 0, 20, 11.36, 15, 5.5]
    [20, 11.6, 0, 11.36, 5, 5.5, 15, 5.5]
    [0, 11.36, 0, 0, 5, 5.5]]  

data = [5, 5.5, 15, 5.5] 
data[0] -> D0
data[1] -> D1

我想用 dict 值和数据数组值替换 values 数组的值。

所以输出应该是；

[
    ["P0", "P1", "D1", "D0"],
    ["P1", "P2", "D1"],
    ["P2", "P3", "D0", "D1"],
    ["P3", "P0", "D0"]
]

我尝试过的是，

for x in range(0,len(values), 1):
    y = 0
    oneD = values[x]

for i, j in valueDict.iteritems():
    print("y ", y)
    print("left : ",(oneD[y], oneD[y+1])," right : ",i)
    if ((oneD[y], oneD[y+1]) == i ):
        oneD[y] = oneD[y].replace(j)
        oneD[y+1] = oneD[y+1].replace(j)

    elif((oneD[y], oneD[y+1]) == data[0]):
        oneD[y] = oneD[y].replace("D0")
        oneD[y+1] = oneD[y+1].replace("D0")

    elif((oneD[y], oneD[y+1]) == data[1]):
        oneD[y] = oneD[y].replace("D1")
        oneD[y+1] = oneD[y+1].replace("D1")

    else:
        y += 2
        continue

    y += 2

此代码无法正常工作。我该怎么做？

【问题讨论】：

标签： python python-3.x dictionary multidimensional-array

【解决方案1】：

代码：

value_dict = {(0, 0): 'P0', (20, 0): 'P1', (20, 11.36): 'P2',
              (0, 11.36): 'P3'}

values = [
    [0, 0, 20, 0, 15, 5.5, 5, 5.5],
    [20, 0, 20, 11.36, 15, 5.5],
    [20, 11.36, 0, 11.36, 5, 5.5, 15, 5.5],
    [0, 11.36, 0, 0, 5, 5.5]
]

data = [5, 5.5, 15, 5.5]

# add data to value_dict
iterator = iter(data)
for i, j in enumerate(iterator):
    value_dict[j, next(iterator)] = 'D{}'.format(i)


# Translate the data
result = []
for v in values:
    line = []
    iterator = iter(v)
    for i in iterator:
        line.append(value_dict[(i, next(iterator))])
    result.append(line)

print(value_dict)
print(result)

结果：

{
    (0, 0): 'P0', 
    (20, 0): 'P1', 
    (20, 11.36): 'P2', 
    (0, 11.36): 'P3', 
    (5, 5.5): 'D0', 
    (15, 5.5): 'D1'
}

[
    ['P0', 'P1', 'D1', 'D0'], 
    ['P1', 'P2', 'D1'], 
    ['P2', 'P3', 'D0', 'D1'], 
    ['P3', 'P0', 'D0']
]

【讨论】：

【解决方案2】：

首先将(5, 5.5): 'D0', (15, 5.5): 'D1'添加到valueDict：

valueDict =  {(0, 0): 'P0', (20, 0): 'P1', (20, 11.36): 'P2', (0, 11.36): 'P3', (5, 5.5): 'D0', (15, 5.5): 'D1'}

并在值中将11.6 更改为11.36：

values = [[0, 0, 20, 0, 15, 5.5, 5, 5.5],
         [20, 0, 20, 11.36, 15, 5.5],
         [20, 11.36, 0, 11.36, 5, 5.5, 15, 5.5],
         [0, 11.36, 0, 0, 5, 5.5]]

首先我们遍历它并将其转换为元组：

tuples_values = []
for v in values:
    tuples_values.append([(v[i], v[i+1]) for i in range(0, len(v), 2)])

然后我们遍历它并用dict的值替换它：

out = []
for v in tuples_values:
    out.append([valueDict[tuple_value] for tuple_value in v])

出来：

[['P0', 'P1', 'D1', 'D0'],
 ['P1', 'P2', 'D1'],
 ['P2', 'P3', 'D0', 'D1'],
 ['P3', 'P0', 'D0']]

【讨论】：