for循环中的concat列

Question

我有以下数据集：

1   8   15  22
2   9   16  23
3   10  17  24
4   11  18  25
5   12  19  26
6   13  20  27
7   14  21  28

我想得到以下结果：

1
2
3
4
5
6
7
8
...
23
24
25
26
27
28

所以我想遍历数据集的所有列，并将每一列连接到第一列。

import pandas as pd

df = pd.read_csv("data.csv", delimiter=";", header=-1)

number_of_columns= len(df.columns)
print(number_of_columns)


for i in range (1,number_of_columns):
  df1 = df.iloc[:,i]
  df2 = pd.concat([df,df1], ignore_index=True)


print(df2)

只有最后一列被连接在最终数据框中。我知道df2在for循环的每次迭代中都会被覆盖。

那我该如何在每个for循环之后“保存” df2，以便每一列都被浓缩？

非常感谢！

Answer 1

逐列显示

stack + tolist

df.stack().tolist()

[1,
 8,
 15,
 22,
 2,
 9,
 16,
 23,
 3,
 10,
 17,
 24,
 4,
 11,
 18,
 25,
 5,
 12,
 19,
 26,
 6,
 13,
 20,
 27,
 7,
 14,
 21,
 28]

逐行

melt

df.melt().value.tolist()

[1,
 2,
 3,
 4,
 5,
 6,
 7,
 8,
 9,
 10,
 11,
 12,
 13,
 14,
 15,
 16,
 17,
 18,
 19,
 20,
 21,
 22,
 23,
 24,
 25,
 26,
 27,
 28]

unstack + tolist

df.unstack().tolist()

#outputs same as above

Answer 2

您也可以这样做：

txt = '''1   8   15  22
2   9   16  23
3   10  17  24
4   11  18  25
5   12  19  26
6   13  20  27
7   14  21  28'''
arr1 = np.fromstring(txt, dtype=int, sep=' ')
arr1.reshape(7,-1).flatten(order = 'F') # for column wise, 'C' can be used for row wise.

Answer 3

您可能不需要第三方库。您可以使用标准库中的csv和itertools模块来返回数字列表：

from io import StringIO
from itertools import chain
import csv

mystr = StringIO("""1   8   15  22
2   9   16  23
3   10  17  24
4   11  18  25
5   12  19  26
6   13  20  27
7   14  21  28""")

with mystr as fin:
    reader = csv.reader(mystr, skipinitialspace=True, delimiter=' ')
    res = list(map(int, chain.from_iterable(zip(*reader))))

print(res)

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10,
 11, 12, 13, 14, 15, 16, 17, 18, 19, 20,
 21, 22, 23, 24, 25, 26, 27, 28]

Answer 4

简单地np.flatten()，

 pd.Series(df.values.flatten())
 (or)
 pd.Series(df.unstack().values)