Python 列中显示的嵌套列表

我正在尝试将嵌套列表显示为列。因此，我正在处理的数据是：

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

我想显示为

  apples Alice  dogs
 oranges   Bob  cats
cherries Carol moose
  banana David goose

使条目右对齐。我已经看过了，但我无法实现类似的结果。到目前为止，我掌握的代码是：

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

total_len= [[] for x in range(len(tableData))]
longest_string = []

for y1 in range(0, len(tableData)):
    for y2 in range(0, len(tableData[y1])):       
        total_len[y1].append(len(tableData[y1][y2]))

for y1 in range(0, len(total_len)):    
    longest_string.append(max(total_len[y1]))

for y1 in range(len(tableData)):
    for y2 in range(len(tableData[y1])):
        print("".join(tableData[y1][y2].rjust(longest_string[y1])))

以及链接线程中的：

>>> for row in zip(*tableData):
...     print("{: >10} {: >10} {: >10}".format(*row))
... 
    apples      Alice       dogs
   oranges        Bob       cats
  cherries      Carol      moose
    banana      David      goose

---

虽然我真的更喜欢。

我不知道你有没有，但如果你有，很容易做到。您可以创建dataframe，然后使用以下方法：

---

没有第三方熊猫的类似格式：

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

# Find the max length of the word in each row
lens = [max(len(col) for col in row) for row in tableData]

# zip(*list) transposes a list...rows become columns
for row in zip(*tableData):
    # Pass the column widths dynamically.
    print('{:>{lens[0]}} {:>{lens[1]}} {:>{lens[2]}}'.format(*row,lens=lens))

输出：

  apples Alice  dogs
 oranges   Bob  cats
cherries Carol moose
  banana David goose

编辑

以下是一个可以动态显示任意数量行和列的版本：

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

# Find the max length of the word in each row
lens = [max(len(col) for col in row) for row in tableData]

# build a format string with an entry for each column
rowfmt = '{:>{}} ' * len(tableData)

# zip(*list) transposes a list...rows become columns
for row in zip(*tableData):
    # Pass the values and column widths dynamically.
    # The zip pairs up each datum with its column width, but in tuples.
    # For example, [data1,data2],[width1,width2] -> [(data1,width1),(data2,width2)]
    # itertools.chain flattens the list of tuples.
    # For example, above becomes [data1,width1,data2,width2]
    print(rowfmt.format(*itertools.chain(*zip(row,lens))))

---

我认为在for the print语句的循环中有3个{:>{lens[I]}形式的条目，因为在这个示例特定的表数据中有3个子列表。是否有可能拥有某种功能，可以获取包含任意数量子列表的任何列表，并以相同的方式显示它？@user49231，是的，需要更多的工作和更复杂的代码。我将添加一个更灵活的示例。

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

# Find the max length of the word in each row
lens = [max(len(col) for col in row) for row in tableData]

# build a format string with an entry for each column
rowfmt = '{:>{}} ' * len(tableData)

# zip(*list) transposes a list...rows become columns
for row in zip(*tableData):
    # Pass the values and column widths dynamically.
    # The zip pairs up each datum with its column width, but in tuples.
    # For example, [data1,data2],[width1,width2] -> [(data1,width1),(data2,width2)]
    # itertools.chain flattens the list of tuples.
    # For example, above becomes [data1,width1,data2,width2]
    print(rowfmt.format(*itertools.chain(*zip(row,lens))))