如何在python中的字典列表中查找项目的累积总和
Posted
技术标签:
【中文标题】如何在python中的字典列表中查找项目的累积总和【英文标题】：How to find cumulative sum of items in a list of dictionaries in python 【发布时间】：2016-12-27 12:31:54 【问题描述】：我有一个类似的列表
a=['time':3,'time':4,'time':5]
我想像这样以相反的顺序获得值的累积总和
b=['exp':3,'cumsum':12,'exp':4,'cumsum':9,'exp':5,'cumsum':5]
最有效的方法是什么？我已经阅读了其他答案，其中使用 numpy 给出了类似
a=[1,2,3]
b=numpy.cumsum(a)
但我还需要在字典中插入 cumsum
【问题讨论】：
最有效的方法是使用 numpy，特别是当列表非常大时。是否需要将 cumsum 放入字典取决于您的应用程序，它可能只是来自糟糕的设计。为什么不使用pandas？它似乎非常适合您的示例。 @ImanolLuengo，谢谢。我的清单不大，但长度不一。此类列表的数量为一百万 使用timeit 或cProfile 来评估解决方案。让我们知道结果。对于将列表转换为另一种对象类型的解决方案，请务必在计时中包含该转换。 【参考方案1】：a=['time':3,'time':4,'time':5]
b = []
cumsum = 0
for e in a[::-1]:
cumsum += e['time']
b.insert(0, 'exp':e['time'], 'cumsum':cumsum)
print(b)
输出：
['exp': 3, 'cumsum': 12, 'exp': 4, 'cumsum': 9, 'exp': 5, 'cumsum': 5]
所以事实证明，在列表开头插入是slow (O(n))。相反，请尝试
deque (O(1))：
from collections import deque
a=['time':3,'time':4,'time':5]
b = deque()
cumsum = 0
for e in a[::-1]:
cumsum += e['time']
b.appendleft('exp':e['time'], 'cumsum':cumsum)
print(b)
print(list(b))
输出：
deque(['cumsum': 12, 'exp': 3, 'cumsum': 9, 'exp': 4, 'cumsum': 5, 'exp': 5])
['cumsum': 12, 'exp': 3, 'cumsum': 9, 'exp': 4, 'cumsum': 5, 'exp': 5]
这是测试 ITT 每种方法的速度的脚本，以及带有时序结果的图表：
from collections import deque
from copy import deepcopy
import numpy as np
import pandas as pd
from random import randint
from time import time
def Nehal_pandas(l):
df = pd.DataFrame(l)
df['cumsum'] = df.ix[::-1, 'time'].cumsum()[::-1]
df.columns = ['exp', 'cumsum']
return df.to_json(orient='records')
def Merlin_pandas(l):
df = pd.DataFrame(l).rename(columns='time':'exp')
df["cumsum"] = df['exp'][::-1].cumsum()
return df.to_dict(orient='records')
def RahulKP_numpy(l):
cumsum_list = np.cumsum([i['time'] for i in l][::-1])[::-1]
for i,j in zip(l,cumsum_list):
i.update('cumsum':j)
def Divakar_pandas(l):
df = pd.DataFrame(l)
df.columns = ['exp']
df['cumsum'] = (df[::-1].cumsum())[::-1]
return df.T.to_dict().values()
def cb_insert_0(l):
b = []
cumsum = 0
for e in l[::-1]:
cumsum += e['time']
b.insert(0, 'exp':e['time'], 'cumsum':cumsum)
return b
def cb_deque(l):
b = deque()
cumsum = 0
for e in l[::-1]:
cumsum += e['time']
b.appendleft('exp':e['time'], 'cumsum':cumsum)
b = list(b)
return b
def cb_deque_noconvert(l):
b = deque()
cumsum = 0
for e in l[::-1]:
cumsum += e['time']
b.appendleft('exp':e['time'], 'cumsum':cumsum)
return b
def hpaulj_gen(l, var='value'):
cum=0
for i in l:
j=i[var]
cum += j
yield var:j, 'sum':cum
def hpaulj_inplace(l, var='time'):
cum = 0
for i in l:
cum += i[var]
i['sum'] = cum
def test(number_of_lists, min_list_length, max_list_length):
test_lists = []
for _ in range(number_of_lists):
test_list = []
number_of_dicts = randint(min_list_length,max_list_length)
for __ in range(number_of_dicts):
random_value = randint(0,50)
test_list.append('time':random_value)
test_lists.append(test_list)
lists = deepcopy(test_lists)
start_time = time()
for l in lists:
res = list(hpaulj_gen(l[::-1], 'time'))[::-1]
elapsed_time = time() - start_time
print('hpaulj generator:'.ljust(25), '%.2f' % (number_of_lists / elapsed_time), 'lists per second')
lists = deepcopy(test_lists)
start_time = time()
for l in lists:
hpaulj_inplace(l[::-1])
elapsed_time = time() - start_time
print('hpaulj in place:'.ljust(25), '%.2f' % (number_of_lists / elapsed_time), 'lists per second')
lists = deepcopy(test_lists)
start_time = time()
for l in lists:
res = cb_insert_0(l)
elapsed_time = time() - start_time
print('craig insert list at 0:'.ljust(25), '%.2f' % (number_of_lists / elapsed_time), 'lists per second')
lists = deepcopy(test_lists)
start_time = time()
for l in lists:
res = cb_deque(l)
elapsed_time = time() - start_time
print('craig deque:'.ljust(25), '%.2f' % (number_of_lists / elapsed_time), 'lists per second')
lists = deepcopy(test_lists)
start_time = time()
for l in lists:
res = cb_deque_noconvert(l)
elapsed_time = time() - start_time
print('craig deque no convert:'.ljust(25), '%.2f' % (number_of_lists / elapsed_time), 'lists per second')
lists = deepcopy(test_lists)
start_time = time()
for l in lists:
RahulKP_numpy(l) # l changed in place
elapsed_time = time() - start_time
print('Rahul K P numpy:'.ljust(25), '%.2f' % (number_of_lists / elapsed_time), 'lists per second')
lists = deepcopy(test_lists)
start_time = time()
for l in lists:
res = Divakar_pandas(l)
elapsed_time = time() - start_time
print('Divakar pandas:'.ljust(25), '%.2f' % (number_of_lists / elapsed_time), 'lists per second')
lists = deepcopy(test_lists)
start_time = time()
for l in lists:
res = Nehal_pandas(l)
elapsed_time = time() - start_time
print('Nehal pandas:'.ljust(25), '%.2f' % (number_of_lists / elapsed_time), 'lists per second')
lists = deepcopy(test_lists)
start_time = time()
for l in lists:
res = Merlin_pandas(l)
elapsed_time = time() - start_time
print('Merlin pandas:'.ljust(25), '%.2f' % (number_of_lists / elapsed_time), 'lists per second')
【讨论】：
谢谢@craig，但这是合理有效的方法吗？因为我有大约一百万个这样的列表 在这里计时各种方法，让我们知道:)【参考方案2】：基于生成器的解决方案：
def foo(a, var='value'):
cum=0
for i in a:
j=i[var]
cum += j
yield var:j, 'sum':cum
In [79]: a=['time':i for i in range(5)]
In [80]: list(foo(a[::-1], var='time'))[::-1]
Out[80]:
['sum': 10, 'time': 0,
'sum': 10, 'time': 1,
'sum': 9, 'time': 2,
'sum': 7, 'time': 3,
'sum': 4, 'time': 4]
在快速测试中，这与 cb_insert_0 具有竞争力
就地版本做得更好：
def foo2(a, var='time'):
cum = 0
for i in a:
cum += i[var]
i['sum'] = cum
foo2(a[::-1])
【讨论】：
Plus 用于 in-place 版本。 添加到时间 :)【参考方案3】：这是使用pandas 的另一种方法-
df = pd.DataFrame(a)
df.columns = ['exp']
df['cumsum'] = (df[::-1].cumsum())[::-1]
out = df.T.to_dict().values()
样本输入、输出-
In [396]: a
Out[396]: ['time': 3, 'time': 4, 'time': 5]
In [397]: out
Out[397]: ['cumsum': 12, 'exp': 3, 'cumsum': 9, 'exp': 4, 'cumsum': 5, 'exp': 5
【讨论】：
pandas 将 dict 转换为 df 并返回的速度有多快？那肯定比cumsum花费更多的时间。 @hpaulj 真的不知道！我猜对一些时间会很好。【参考方案4】：试试这个：
a = ['time':3,'time':4,'time':5]
df = pd.DataFrame(a).rename(columns='time':'exp')
df["cumsum"] = df['exp'][::-1].cumsum()
df.to_dict(orient='records')
字典没有顺序。
['cumsum': 12, 'exp': 3, 'cumsum': 9, 'exp': 4, 'cumsum': 5, 'exp': 5]
【讨论】：
【参考方案5】：试试这个，
cumsum_list = np.cumsum([i['time'] for i in a][::-1])[::-1]
for i,j in zip(a,cumsum_list):
i.update('cumsum':j)
结果
['cumsum': 12, 'time': 3, 'cumsum': 9, 'time': 4, 'cumsum': 5, 'time': 5]
效率
改成函数，
In [49]: def convert_dict(a):
....: cumsum_list = np.cumsum([i['time'] for i in a][::-1])[::-1]
....: for i,j in zip(a,cumsum_list):
....: i.update('cumsum':j)
....: return a
然后是结果，
In [51]: convert_dict(a)
Out[51]: ['cumsum': 12, 'time': 3, 'cumsum': 9, 'time': 4, 'cumsum': 5, 'time': 5]
最后是效率，
In [52]: %timeit convert_dict(a)
The slowest run took 12.84 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 12.1 µs per loop
【讨论】：
【参考方案6】：使用pandas:
In [4]: df = pd.DataFrame(['time':3,'time':4,'time':5])
In [5]: df
Out[5]:
time
0 3
1 4
2 5
In [6]: df['cumsum'] = df.ix[::-1, 'time'].cumsum()[::-1]
In [7]: df
Out[7]:
time cumsum
0 3 12
1 4 9
2 5 5
In [8]: df.columns = ['exp', 'cumsum']
In [9]: df
Out[9]:
exp cumsum
0 3 12
1 4 9
2 5 5
In [10]: df.to_json(orient='records')
Out[10]: '["exp":3,"cumsum":12,"exp":4,"cumsum":9,"exp":5,"cumsum":5]'
【讨论】：
以上是关于如何在python中的字典列表中查找项目的累积总和的主要内容，如果未能解决你的问题，请参考以下文章
使用带有 ORDER BY 的 SQL Server 查找累积总和