我很难在我的函数中实现移动平均公式。 我花了相当长的一段时间才能了解代码的正确位置。
有没有可能我可以租用的图书馆?
输入:
ma([2,3,4,3,2,6,9,3,2,1], 4)
预期输出:
[None, None, None, 3.0, 3.0, 3.75, 5.0, 5.0, 5.0, 3.75]
我的输出:
[None, None, 0.0, 3.0, 3.75, 5.0, 5.0, None, None, None]
我遇到的问题是,结果的中间部分是正确的,但其余部分仍然是个谜。
def ma(prices,n):
ma = []
sums = []
s = 0
ave = 0
for idx, i in enumerate(prices):
s += i
sums.append(s)
print('idx: ' + str(idx))
print('list of sums ' + str(sums))
#print('sum ' + str(s))
if i >= n+1:
print('sums[idx] ' + str(sums[idx]))
print('sums[idx-n] ' + str(sums[idx-n]))
ave = (sums[idx] - sums[idx-n]) / n
print('ave ' + str(ave))
ma.append(ave)
print('ma ' + str(ma))
else:
m = None
ma.append(m)
print('ma ' + str(ma))
(对所有这些print
函数调用都很抱歉,但是我真的很想找到问题的根源)。
答案 0 :(得分:2)
您的代码中还有其他逻辑错误。我试图对其进行纠正以使其按需工作。以下仅是for循环的修改版本。休息保持不变。添加/修改的行以注释突出显示
for idx, i in enumerate(prices):
s += i
sums.append(s)
if idx == n-1: # Added
ave = (sums[idx]) / n # Added
ma.append(ave) # Added
elif idx >= n: # modified
ave = (sums[idx] - sums[idx-n]) / n
ma.append(ave)
else:
ma.append(None) # removed extra variable m
问题是您使用了错误的变量作为索引:
一个主要问题是您正在使用
if i >= n+1:
您应使用:
if idx >= n+1:
此外,我添加了一个if语句来处理前三个元素的平均值。
现在
moving_average([2,3,4,5,8,5,4,3,2,1], 3)
提供以下输出(您可以稍后舍入):
[None, None, 3.0, 4.0, 5.666666666666667, 6.0, 5.666666666666667, 4.0, 3.0, 2.0]
答案 1 :(得分:1)
如果可以使用标准库,则可能会有所帮助。您真正需要的是迭代器上的滑动窗口。您可以为此使用此功能(该功能基于itertools食谱中的grouper
):
from itertools import islice
def window(iterable, n=2):
# window('123', 2) --> '12' '23'
args = [islice(iterable, i, None) for i in range(n)]
return zip(*args)
对于平均值,您可以使用statistics.mean
。 paddig部分可以简单地通过将平均列表与[None] * (n - 1)
相加来实现:
from statistics import mean
def moving_average(prices, n):
avgs = [mean(w) for w in window(prices, n)]
padding = [None] * (n - 1)
return padding + avgs
样品用量:
>>> moving_average([2,3,4,5,8,5,4,3,2,1], 3)
[None, None, 3, 4, 5.666666666666667, 6, 5.666666666666667, 4, 3, 2]
>>> moving_average([1, 2, 3], 3)
[None, None, 2]
>>> moving_average([1, 2, 3], 1)
[1, 2, 3]
>>> moving_average([5, 10, 0], 2)
[None, 7.5, 5]
答案 2 :(得分:0)
程序返回stateToProps
的原因是负索引。当var INPUT = 'INPUT';
var setInput = (val = '') => ({
type: INPUT,
payload: val
})
var modelA = (state = {test1: '', test2: false, inputVal: ''}, action) => {
switch (action.type) {
case this.INPUT:
state.inputVal = action.payload;
return state
default:
return state
}
}
var modelB = (state = {test1: '', content: ''}, action) => {
switch (action.type) {
default:
return state
}
}
var stateToProps = state => {
return {
storeValue: state.modelA.inputVal
}
}
var dispatchToProps = dispatch => {
return {
setStoreVal: (val) => {
dispatch(setInput(val))
}
}
}
class App extends React.Component {
constructor(props) {
super(props);
this.store = Redux.createStore(Redux.combineReducers({modelA, modelB}));
}
render() {
return e(
ReactRedux.Provider, {store:this.store},
e('div', null, "App",
e(ReactRedux.connect(stateToProps, dispatchToProps)(Whatever), null)
)
);
}
}
class Whatever extends React.Component {
constructor(props) {
super(props);
this.state = {val:0}
this.listener = this.listener.bind(this);
}
listener(evt) {
this.props.setStoreVal(evt.currentTarget.value);
this.setState({val:evt.currentTarget.value});
}
render() {
return e(
'div', null,
e('p', null, this.props.storeValue),
e('input', {onChange:this.listener, val:this.state.val})
);
}
}
为9-9 / 3 = 0
时,idx
在说2
,它指向列表的最后一项,sums[idx-n]
。 Understanding Python's slice notation可以帮助解释这一点。
答案 3 :(得分:0)
您还可以使用列表切片来解决此问题,以对输入列表进行智能分区并计算列表分区的平均值:
def moving_average(data,window):
"""The partitions begin with window-1 None. Then follow partial lists, containing
window-sized elements. We do this only up to len(data)-window+1 as the following
partitions would have less then window elements."""
parts = [None]*(window-1) + [ data[i:i+window] for i in range(len(data)-window+1)]
# The None's The sliding window of window elements
# we return None if the value is None else we calc the avg
return [ sum(x)/window if x else None for x in parts]
print( moving_average([2,3,4,5,8,5,4,3,2,1], 1) )
print( moving_average([2,3,4,5,8,5,4,3,2,1], 2) )
print( moving_average([2,3,4,5,8,5,4,3,2,1], 3) )
输出(包括parts
作为注释):
# [[2], [3], [4], [5], [8], [5], [4], [3], [2], [1]]
[2.0, 3.0, 4.0, 5.0, 8.0, 5.0, 4.0, 3.0, 2.0, 1.0]
# [None, [2, 3], [3, 4], [4, 5], [5, 8], [8, 5], [5, 4], [4, 3], [3, 2], [2, 1]]
[None, 2.5, 3.5, 4.5, 6.5, 6.5, 4.5, 3.5, 2.5, 1.5]
# [None, None, [2, 3, 4], [3, 4, 5], [4, 5, 8], [5, 8, 5], [8, 5, 4],
# [5, 4, 3], [4, 3, 2], [3, 2, 1]]
[None, None, 3.0, 4.0, 5.666666666666667, 6.0, 5.666666666666667, 4.0, 3.0, 2.0]