Question

我有两个数组。 column_names保留列标题。 values保留所有值。

我了解是否可以这样做

column_names = ["a", "b", "c"]
values = [1, 2, 3]
for n, v in zip(column_names, values):
    print("{} = {}".format(n, v))

我明白了

a = 1
b = 2
c = 3

如果通过验证，我该如何编码：

column_names = ["a", "b", "c"]
values = [1, 2, 3, 4, 5, 6, 7, 8, 9]

我会得到

a = 1, 4, 7
b = 2, 5, 8
c = 3, 6, 9

谢谢！

Answer 1

您可以执行以下操作

>>> for n, v in zip(column_names, zip(*[values[i:i+3] for i in range(0,len(values),3)])):
...     print("{} = {}".format(n, ', '.join(map(str, v))))
... 
a = 1, 4, 7
b = 2, 5, 8
c = 3, 6, 9

或者，您可以使用在grouper中定义的itertools

>>> def grouper(iterable, n, fillvalue=None):
...     "Collect data into fixed-length chunks or blocks"
...     # grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx"
...     args = [iter(iterable)] * n
...     return zip_longest(*args, fillvalue=fillvalue)
... 
>>> from itertools import zip_longest
>>> for n, v in zip(column_names, zip(*grouper(values, 3))):
...     print("{} = {}".format(n, ', '.join(map(str, v))))
... 
a = 1, 4, 7
b = 2, 5, 8
c = 3, 6, 9

Answer 2

使用pandas和numpy很容易，结果将是一个更有用的表。熊猫擅长安排表格数据。因此，让我们利用它：用以下命令安装熊猫：

pip install pandas

#pandas comes with numpy
import numpy as np
import pandas as pd

# this makes a normal python list for integers 1-9
input = list(range(1,10))

#lets convert that to numpy array
num = numpy.array(input)

#currently its shape is single dimensional, lets change that to a two dimensional matrix that turns it into the clean breaks you want
reshaped = num.reshape(3,3)

#now construct a beautiful table
pd.DataFrame(reshaped, columns=['a','b','c'])

#ouput is
   a  b  c
0  1  2  3
1  4  5  6
2  7  8  9

Answer 3

在这种情况下，

itertools.cycle似乎合适。这是供将来的读者使用的另一个版本：

import itertools

column_names = ["a", "b", "c"]
values = [1, 2, 3, 4, 5, 6, 7, 8, 9]

L = zip(itertools.cycle(column_names), values)

for g, v in itertools.groupby(sorted(L), lambda x: x[0]):
    print("{} = {}".format(g, [i[1] for i in v]))

给予：

a = [1, 4, 7]
b = [2, 5, 8]
c = [3, 6, 9]

Answer 4

您需要执行两个子步骤。

首先，您希望将列表分成多个块，然后再将这些块分配给字典。

要将列表分成多个部分，我们可以创建一个函数：

Functor

这时我们将有： [[1,4,7]，[2,5,8]，[3,6,9]]

从这里开始，创建dict非常简单。首先，我们将两个列表压缩在一起：

def chunk(values, chunk_size):
    assert len(values)%chunk_size == 0 # Our chunk size has to evenly fit in our list
    steps = len(values)/chunk_size
    chunky_list = []
    for i in range(0,steps):
        position = 0 + i
        sub_list = []
        while position < len(values):
            sub_list.append(values[position])
            position += chunk_size
        chunky_list.append(sub_list)
    return chunky_list

利用Python知道如何将元组列表转换为字典这一事实：

zip(column_names, chunk(3))

Answer 5

您还可以使用切片和collections.defaultdict来收集您的值：

from collections import defaultdict

column_names = ["a", "b", "c"]
values = [1, 2, 3, 4, 5, 6, 7, 8, 9]

column_len = len(column_names)

d = defaultdict(list)
for i in range(0, len(values), column_len):
    seq = values[i:i+column_len]

    for idx, number in enumerate(seq):
        d[column_names[idx]].append(number)

for k, v in d.items():
    print('%s = %s' % (k, ', '.join(map(str, v))))

哪些输出：

a = 1, 4, 7
b = 2, 5, 8
c = 3, 6, 9

如果您使用itertools.cycle创建压缩列表，而避免同时切片，则可以避免这种情况：

from collections import defaultdict
from itertools import cycle

column_names = ["a", "b", "c"]
values = [1, 2, 3, 4, 5, 6, 7, 8, 9]

column_names = cycle(column_names)

d = defaultdict(list)
for column, val in zip(column_names, values):
    d[column].append(val)

for k, v in d.items():
    print('%s = %s' % (k, ', '.join(map(str, v))))

将列表转换成表格-Python

5 个答案: