Question

说我有一个列表：

nlist = [0,0,0,0, 0,0,0,0, 0,0,0,0, 0,0,0,0]

我有一些代码：

def drawBoard():
    count = 0
    for i in range(16):
        print('|',nlist[i], end = ' ')
        count += 1
        if count == 4:
            print("|\n-----------------")
            count = 0
    print("")

此代码显示如下列表：

| 0 | 0 | 0 | 0 |
-----------------
| 0 | 0 | 0 | 0 |
-----------------
| 0 | 0 | 0 | 0 |
-----------------
| 0 | 0 | 0 | 0 |
-----------------

现在板子看起来不错，但是说我有一个像128这样的值，而不是零。那样会使木板错位，并使它看起来有些混乱，但是当我得到一堆数字时，木板看起来会很混乱，像这样：

| 128 | 0 | 4096 | 0 |
-----------------
| 64 | 0 | 0 | 128 |
-----------------
| 2048 | 0 | 0 | 8192 |
-----------------
| 0 | 0 | 64 | 0 |
-----------------

现在板子看起来一团糟，每个数字下面的行都没到尽头。我怎样才能使电路板看起来像这样（也许改变直线之间的间距），以使它不再显得凌乱：

| 128  | 0 | 4096 |   0  |
--------------------------
|  64  | 0 |  0   | 128  |
--------------------------
| 2048 | 0 |   0  | 4096 |
--------------------------
|  0   | 0 |  64  |  0   |
--------------------------

Answer 1

下面的代码在列表中找到最大数字的长度，然后在每个数字的两边分别添加填充。之后，我们通过遍历列表4个步骤来打印每一行。

在：

L = [128,0,4096,0,64,0,0,128,2048,0,0,4096,0,0,64,0]

m = len(str(max(L))) + 2
L = [ str(x).center(m,' ') for x in L]
step = 4

for i in range(0,len(L),step):
    s = '|' + '|'.join(L[i:i+step]) + '|'
    print(s)
    print('-'*len(s))

出局：

| 128  |  0   | 4096 |  0   |
-----------------------------
|  64  |  0   |  0   | 128  |
-----------------------------
| 2048 |  0   |  0   | 4096 |
-----------------------------
|  0   |  0   |  64  |  0   |
-----------------------------

如果您想要更大或更小的表，只需更改步骤的值即可。

在：

L = [128,0,4096,0,64,0,0,128,2048,0,0,4096,0,0,64,0]

m = len(str(max(L))) + 2
L = [ str(x).center(m,' ') for x in L]
step = 8

for i in range(0,len(L),step):
    s = '|' + '|'.join(L[i:i+step]) + '|'
    print(s)
    print('-'*len(s))

出局：

| 128  |  0   | 4096 |  0   |  64  |  0   |  0   | 128  |
---------------------------------------------------------
| 2048 |  0   |  0   | 4096 |  0   |  0   |  64  |  0   |
---------------------------------------------------------

但是，如果您希望每列的填充都尽可能紧而又不显得难看，那么事情会变得更加复杂。

在：

L = [128,0,4096,0,64,0,0,128,2048,0,0,4096,0,0,64,0]
step = 4

# We calculate the padding for every column and store it in sz[].
# The padding for a column is the same as the minimum padding for its largest element.
sz = []
for index,item in enumerate(L):
    column = [L[i] for i in range(index,len(L),step)]
    padding = len( str( max(column) ) ) + 2
    sz.append(padding)
    if index == step + 1:
        break

# Now we iterate over the columns ( not linearly over the array ) 
# and we apply the appropriate padding we have just found out.
i = 0
while i < step + 1:
    for j in range(i,len(L),step):
        L[j] = str(L[j]).center(sz[i],' ')
    i += 1

# Now simply iterate linearly over the array and print a new line 
# every step.
for i in range(0,len(L),step):
    s = '|' + '|'.join(L[i:i+step]) + '|'
    print(s)
    print('-'*len(s))

出局：

| 128  | 0 | 4096 |  0   |
--------------------------
|  64  | 0 |  0   | 128  |
--------------------------
| 2048 | 0 |  0   | 4096 |
--------------------------
|  0   | 0 |  64  |  0   |
--------------------------

Answer 2

首先计算每列的最大宽度，然后使用字符串格式化程序以该列的最大宽度填充每个数字，并使其居中对齐。行分隔符的宽度就是所有列宽加上5个列分隔符的总和。

nlist = [128,0,4096,0, 64,0,0,128, 2048,0,0,8192, 0,0,64,0]
def drawBoard():
    widths = [max(len(str(nlist[row * 4 + col])) for row in range(4)) + 2 for col in range(4)]
    width = sum(widths) + 5
    count = 0
    for i in range(16):
        print('|{:^{width}}'.format(nlist[i], width=widths[i % 4]), end = '')
        count += 1
        if count == 4:
            print("|\n" + '-' * width)
            count = 0
    print("")
drawBoard()

这将输出：

| 128  | 0 | 4096 |  0   |
--------------------------
|  64  | 0 |  0   | 128  |
--------------------------
| 2048 | 0 |  0   | 8192 |
--------------------------
|  0   | 0 |  64  |  0   |
--------------------------

使2 x 2板居中

2 个答案: