[英]Split rows into multiple rows with pandas
I have a dataset in the following format. 我有以下格式的数据集。 It got 48 columns and about 200000 rows.
它有48列和大约200000行。
slot1,slot2,slot3,slot4,slot5,slot6...,slot45,slot46,slot47,slot48
1,2,3,4,5,6,7,......,45,46,47,48
3.5,5.2,2,5.6,...............
I want to reshape this dataset to something as below, where N is less than 48 (maybe 24 or 12 etc..) column headers doesn't matter. 我想将此数据集重塑为以下内容,其中N小于48(也许是24或12等)列标题无关紧要。 when N = 4
当N = 4时
slotNew1,slotNew2,slotNew3,slotNew4
1,2,3,4
5,6,7,8
......
45,46,47,48
3.5,5.2,2,5.6
............
I can read row by row and then split each row and append to a new dataframe. 我可以逐行阅读,然后将每一行拆分并追加到新的数据框中。 But that is very inefficient.
但这是非常低效的。 Is there any efficient and faster way to do that?
有什么有效,快捷的方法可以做到这一点吗?
You may try this 你可以试试这个
N = 4
df_new = pd.DataFrame(df_original.values.reshape(-1, N))
df_new.columns = ['slotNew{:}'.format(i + 1) for i in range(N)]
The code extracts the data into numpy.ndarray
, reshape it, and create a new dataset of desired dimension. 该代码将数据提取到
numpy.ndarray
,对其进行numpy.ndarray
,并创建一个具有所需尺寸的新数据集。
Example: 例:
import numpy as np
import pandas as pd
df0 = pd.DataFrame(np.arange(48 * 3).reshape(-1, 48))
df0.columns = ['slot{:}'.format(i + 1) for i in range(48)]
print(df0)
# slot1 slot2 slot3 slot4 ... slot45 slot46 slot47 slot48
# 0 0 1 2 3 ... 44 45 46 47
# 1 48 49 50 51 ... 92 93 94 95
# 2 96 97 98 99 ... 140 141 142 143
#
# [3 rows x 48 columns]
N = 4
df = pd.DataFrame(df0.values.reshape(-1, N))
df.columns = ['slotNew{:}'.format(i + 1) for i in range(N)]
print(df.head())
# slotNew1 slotNew2 slotNew3 slotNew4
# 0 0 1 2 3
# 1 4 5 6 7
# 2 8 9 10 11
# 3 12 13 14 15
# 4 16 17 18 19
Another approach 另一种方法
N = 4
df1 = df0.stack().reset_index()
df1['i'] = df1['level_1'].str.replace('slot', '').astype(int) // N
df1['j'] = df1['level_1'].str.replace('slot', '').astype(int) % N
df1['i'] -= (df1['j'] == 0) - df1['level_0'] * 48 / N
df1['j'] += (df1['j'] == 0) * N
df1['j'] = 'slotNew' + df1['j'].astype(str)
df1 = df1[['i', 'j', 0]]
df = df1.pivot(index='i', columns='j', values=0)
Use pandas.explode
after making chunks. 制作大块后使用
pandas.explode
。 Given df
: 给定
df
:
import pandas as pd
df = pd.DataFrame([np.arange(1, 49)], columns=['slot%s' % i for i in range(1, 49)])
print(df)
slot1 slot2 slot3 slot4 slot5 slot6 slot7 slot8 slot9 slot10 ... \
0 1 2 3 4 5 6 7 8 9 10 ...
slot39 slot40 slot41 slot42 slot43 slot44 slot45 slot46 slot47 \
0 39 40 41 42 43 44 45 46 47
slot48
0 48
Using chunks
to divide: 使用
chunks
划分:
def chunks(l, n):
"""Yield successive n-sized chunks from l.
Source: https://stackoverflow.com/questions/312443/how-do-you-split-a-list-into-evenly-sized-chunks
"""
n_items = len(l)
if n_items % n:
n_pads = n - n_items % n
else:
n_pads = 0
l = l + [np.nan for _ in range(n_pads)]
for i in range(0, len(l), n):
yield l[i:i + n]
N = 4
new_df = pd.DataFrame(list(df.apply(lambda x: list(chunks(list(x), N)), 1).explode()))
print(new_df)
Output: 输出:
0 1 2 3
0 1 2 3 4
1 5 6 7 8
2 9 10 11 12
3 13 14 15 16
4 17 18 19 20
...
Advantage of this approach over numpy.reshape
is that it can handle when N
is not a factor: 这种方法优于
numpy.reshape
优势在于,它可以在N
不是一个因素时进行处理:
N = 7
new_df = pd.DataFrame(list(df.apply(lambda x: list(chunks(list(x), N)), 1).explode()))
print(new_df)
Output: 输出:
0 1 2 3 4 5 6
0 1 2 3 4 5 6 7.0
1 8 9 10 11 12 13 14.0
2 15 16 17 18 19 20 21.0
3 22 23 24 25 26 27 28.0
4 29 30 31 32 33 34 35.0
5 36 37 38 39 40 41 42.0
6 43 44 45 46 47 48 NaN
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