我有一个包含1000多列的数据帧,并且我有预定义的组列表。我想将每个单元格值和每个组边界进行比较,并创建一个新列来分配组名称。我已经写了for loops
,但它花了5分钟多的时间来处理。有没有更有效的方法来实现这一点?感谢
这是我的数据帧
Frequency
21.0
18.0
16.0
10.0
10.0
9.0
10.0
10.0
5.0
8.0
和我预定义的群组列表
> groups
[(3, 5), (6, 10), (11, 30)]
我想要的是
Frequency Group
21.0 11-30
18.0 11-30
16.0 11-30
10.0 6-10
10.0 6-10
9.0 6-10
10.0 6-10
10.0 6-10
5.0 3-5
8.0 6-10
这是我的代码
for i in range(0, len(fre_table["Frequency"])):
for j in range(0, len(groups)):
if fre_table["Frequency"][i] >= groups[j][0] and fre_table["Frequency"][i] <= groups[j][1]:
break
fre_table['Group'][i] = "{}-{}".format(groups[j][0], groups[j][1])
确定@BallpointBen在评论部分中建议的解决方案的效率
数据:
import numpy as np
import pandas as pd
fre_table = pd.DataFrame({'Index':[0,1,2,3,4,5,6,7,8,9],
'Frequency':[21.0, 18.0, 16.0, 10.0, 10.0, 9.0, 10.0, 10.0, 5.0, 8.0]})
groups = [(3, 5), (6, 10), (11, 30)]
初始解决方案所花费的时间:0.5420
import timeit
start_time = timeit.default_timer()
fre_table['Group'] = 0
for i in range(0, len(fre_table["Frequency"])):
for j in range(0, len(groups)):
if fre_table["Frequency"][i] >= groups[j][0] and fre_table["Frequency"][i] <= groups[j][1]:
break
fre_table['Group'][i] = "{}-{}".format(groups[j][0], groups[j][1])
elapsed_time = timeit.default_timer() - start_time
最终解决方案所需时间:0.0043s
import timeit
start_time = timeit.default_timer()
bins = pd.IntervalIndex.from_tuples(groups)
fre_table['Group'] = pd.cut(fre_table['Frequency'], bins)
elapsed_time = timeit.default_timer() - start_time
大约快100倍!