Question

我在某些条件下制作名单。

这就是它的样子。

def time_price_pair(a, b):
    if 32400<=a and a<32940:
        a_list=[]
        a_list.append(b)
    elif 32940<=a and a<33480:
        b_list=[]
        b_list.append(b)
    elif 33480<=a and a <34020:
        c_list=[]
        c_list.append(b)
    ......
    ......
    ......
    elif 52920 <=a and a <53460:
        some_list=[]
        some_list.append(b)

每个条件将由540添加。如[32400,32940,33480,34020,34560,35100,35640,36180,36720,37260,37800,38340,38880,39420 .... 53460]

和列表名称并不重要。

Answer 1

您可以使用带有一些递增变量i的for循环，并不断更新要求。像这样：

def time_price_pair(a, b):
    min = 32400
    max = 32940
    inc = 540
    for i in range(some value):
        if min + inc*i <= a < max + inc*i:
            b = min + inc*i
            a_list = [b]

Answer 2

我会使用dict存储这些值列表，并使用一些数学知道将这些数字放在哪里

from collections import defaultdict
lists = defaultdict(list)

def time_price_pair(a, b):
    if 32400 <= a < 53460:
        i = (a-32400)/540
        lists[i].append(b)

Answer 3

字典可用于保存所有使用的时间范围箱，如下所示：

import collections

time_prices = [(32401, 20), (32402,30), (32939, 42), (32940, 10), (32941, 15), (40000, 123), (40100, 234)]
dPrices = collections.OrderedDict()

for atime, aprice in time_prices:
    abin = 32400 + ((atime - 32400) // 540) * 540    # For bins as times
    #abin = (atime - 32400) // 540 + 1               # For bins starting from 1
    dPrices.setdefault(abin, []).append(aprice)

# Display results
for atime, prices in dPrices.items():
    print atime, prices

这会给你以下输出：

32400 [20, 30, 42]
32940 [10, 15]
39960 [123, 234]

或单独作为：

print dPrices[32400]
[20, 30, 42]

使用Python 2.7进行测试

Answer 4

看起来简单的高级熊猫函数pd.cut非常适合您的目的。

import pandas as np
import numpy as np

# simulate your data
# ==================================
np.random.seed(0)
a = np.random.randint(32400, 53439, size=1000000)
b = np.random.randn(1000000)

# put them in dataframe
df = pd.DataFrame(dict(a=a, b=b))
print(df)


            a       b
0       35132 -0.4605
1       43199 -0.9469
2       42245  0.2580
3       52048 -0.7309
4       45523 -0.4334
5       41625  2.0155
6       53157 -1.4712
7       46516 -0.1715
8       47335 -0.6594
9       47830 -1.0391
...       ...     ...
999990  39754  0.8771
999991  34779  0.7030
999992  37836  0.5409
999993  44330 -0.6747
999994  41078 -1.1368
999995  38752  1.6121
999996  42155 -0.1139
999997  49018 -0.1737
999998  45848 -1.2640
999999  50669 -0.4367

# processing
# ===================================
rng = np.arange(32400, 53461, 540)
# your custom labels
labels = np.arange(1, len(rng))
# use pd.cut()
%time df['cat'] = pd.cut(df.a, bins=rng, right=False, labels=labels)

CPU times: user 52.5 ms, sys: 16 µs, total: 52.5 ms 
Wall time: 51.6 ms

print(df)

            a       b cat
0       35132 -0.4605   6
1       43199 -0.9469  20
2       42245  0.2580  19
3       52048 -0.7309  37
4       45523 -0.4334  25
5       41625  2.0155  18
6       53157 -1.4712  39
7       46516 -0.1715  27
8       47335 -0.6594  28
9       47830 -1.0391  29
...       ...     ...  ..
999990  39754  0.8771  14
999991  34779  0.7030   5
999992  37836  0.5409  11
999993  44330 -0.6747  23
999994  41078 -1.1368  17
999995  38752  1.6121  12
999996  42155 -0.1139  19
999997  49018 -0.1737  31
999998  45848 -1.2640  25
999999  50669 -0.4367  34

[1000000 rows x 3 columns]


# groupby
grouped = df.groupby('cat')['b']

# access to a particular group using your user_defined key
grouped.get_group(1).values

array([ 0.4525, -0.7226, -0.981 , ...,  0.0985, -1.4286, -0.2257])

我怎样才能缩短这个python代码？

4 个答案: