Question

我关注http://automatetheboringstuff.com/chapter6/
在页面的最底部是关于格式化表格的练习。

这是我的代码：

tableData = [['apples', 'oranges', 'cherries', 'banana'],
         ['Alice', 'Bob', 'Carol', 'David'],
         ['dogs', 'cats', 'moose', 'goose']]

def printTable(table):
    colWidths = [0] * len(table)

    for line in table:
        max = 0
        for word in line:
            if len(word) > max:
                max = len(word)
        colWidths[table.index(line)] = max

    for a in range(len(table)-2):
        for b in range(len(table[0])):
            print(table[a][b].rjust(colWidths[0])+table[a+1][b].rjust(colWidths[1])+table[a+2][b].rjust(colWidths[2]))

    """
    print(table[0][0].rjust(colWidths[0]), table[1][0].rjust(colWidths[1]), table[2][0].rjust(colWidths[2]))
    print(table[0][1].rjust(colWidths[0]), table[1][1].rjust(colWidths[1]), table[2][1].rjust(colWidths[2]))
    print(table[0][2].rjust(colWidths[0]), table[1][2].rjust(colWidths[1]), table[2][2].rjust(colWidths[2]))
    print(table[0][3].rjust(colWidths[0]), table[1][3].rjust(colWidths[1]), table[2][3].rjust(colWidths[2]))
    """
    print()

printTable(tableData)

注释掉的行格式化了应有的一切。实际的代码没有。为了正确格式化，我需要为每列添加1到.rjust()（例如，我需要100列.rjust(colWidths[1]+99)）。

当我手动打印时，为什么会出现这种情况呢？

Answer 1

for循环中的print语句使用字符串连接：

print(table[a][b].rjust(colWidths[0])+table[a+1][b].rjust(colWidths[1])+table[a+2][b].rjust(colWidths[2]))

连接在内存中创建每个字符串，然后在新字符串中将它们组合在一起。不会在项目之间添加空格，这就是为什么你需要在它之前为每一列添加一个字符。

代码中注释掉的行使用逗号分隔参数：

"""
print(table[0][0].rjust(colWidths[0]), table[1][0].rjust(colWidths[1]), table[2][0].rjust(colWidths[2]))
print(table[0][1].rjust(colWidths[0]), table[1][1].rjust(colWidths[1]), table[2][1].rjust(colWidths[2]))
print(table[0][2].rjust(colWidths[0]), table[1][2].rjust(colWidths[1]), table[2][2].rjust(colWidths[2]))
print(table[0][3].rjust(colWidths[0]), table[1][3].rjust(colWidths[1]), table[2][3].rjust(colWidths[2]))
"""

带有逗号分隔项的print语句，使用空格分隔它们。这可能是您的列正确排列的原因。

This answer更详细地解释了它。

Answer 2

你所挣扎的代码部分与我和我的工作非常相似。

def printTable(List):
    colWidths = [0] * len(tableData)

    for line in range(len(tableData)):
        for word in range(len(tableData[line])):
            if colWidths[line] <= len(tableData[line][word]):
                colWidths[line] = len(tableData[line][word])
            else:
                colWidths[line] = colWidths[line]

#this is the part where you struggled
    for li in range(len(tableData[0])):
        for i in range(len(tableData)):
            print(tableData[i][li].rjust(colWidths[i]), end =" ")
        print()

Answer 3

下面是我用来在表格中找到最长字符串并打印表格的代码

(click)="..."

def printTable（table）：

tableData = [['apples', 'orange', 'cherries', 'banana'],
         ['Alice', 'Bob', 'Carol', 'David'],
         ['dogs', 'cats', 'moose', 'goose']]

Answer 4

由于每个单独列表的内容都定义了每列必需的宽度，因此可以在循环中执行len（max（tableData [x]））以获得每列的最大长度。将此附加到列表中，即可轻松传输：

tableData = [['apples', 'oranges', 'cherries', 'banana'],
             ['Alice', 'Bob', 'Carol', 'David'],
             ['dogs', 'cats', 'moose', 'goose']]

colWidth = []

def printTable(table):
    for x in range(len(table)):
        colWidth.append(len(max(table[x],key=len))+1)       

    for x in range(len(table[0])):
        for i in range(len(table)):
            print (table [i][x].rjust(colWidth[i]), end = " ")
        print()

printTable(tableData)

Answer 5

这是我的代码，经过一周的烦扰并追逐着我的梦，终于可以正常工作了。

tableData = [['apples', 'oranges', 'cherries', 'banana'],
             ['Alice', 'Bobolon', 'Carolina', 'Davidovv'],
             ['dogsee', 'puscats', 'moosara', 'gooseano']]

def printTable(table):
    colwidths=[0]*len(table)
    for i in range(len(table)):
        for j in range(len(table[i])):
            if len(table[i][j])>colwidths[i]:
                colwidths[i]=len(table[i][j])

    for line in range(len(table[0])): #PYTHON PRINTS LINE PER LINE. NOT COLUMN PER COLUMN.ABOUT TABLE[0]: IT IS ESTABLISHED THAT ALL ITEMS IN TABLEDATA WILL HVE THE SAME LENGTH, SO THAT IT DOESNT MATTER WETHER YOU PUT [0], [1] OR [2] BECAUSE THEY ALL HAVE LENGTH OF 4 ITEMS (IN CASE OF THE TABLEDATA LIST
        for column in range(len(table)): #THERE ARE AS MANY COLUMNS AS ITEMS(LISTS) IN TABLEDATA
            print(table[column][line].ljust(colwidths[column]*2),end=" ") #NOW, WE PRINT THE FIRST WORD OF THE FIRST COLUMN, FOLLOWED BUYT THE FIRST WORD OF THE SECOND COLUMN AND SO ON. .END= HELPS WITH NOT HAVING TO CONCATENATE THESE AND KEEPING ITEMS IN THE SAME LINE.
        print() #WITHOUT THIS PRINT (WHICH PRINTS A NEW LINE), ALL ITEMS WOULD BE IN THE SAME LINE, DUE TO THE PREVIOS .END=

printTable(tableData)

Answer 6

tableData = [['apples', 'oranges', 'cherries', 'bananas'],
            ['Alice', 'Bob', 'Caroline', 'David'],
            ['dogs', 'cats', 'moose', 'goose']]
#To calculate the length of the longest word in the table
colwid = 0
for j in range(len(tableData[0])):
    for i in range(len(tableData)):
        if colwid < len(tableData[i][j]):
            colwid = len(tableData[i][j])
        i = i + 1
    j = j + 1
#Print the table with each field left justified with column length from above
for j in range(len(tableData[0])):
    for i in range(len(tableData)):
        print(tableData[i][j].ljust(colwid), end=' ')
        i = i + 1
    j = j + 1
    print()

自动化无聊的东西，第6章表打印机

6 个答案: