我有一個(gè)列表，我想計(jì)算它的平均值。獲得平均值后，我想從列表中減去每個(gè)項(xiàng)目的平均值。獲得所有值后，我想平方并求和。x=[17,13,12,15,16,14,16,16,18,19]Average is 15.6For example list look like: x=[17,13,12,15,16,14,16,16,18,19]第1步：Find average valuesum(x)/len(x) average value is 15.6第 2 步：從列表中減去每個(gè)項(xiàng)目的平均值17-15.6 = 1.413-15.6 = -2.612-15.6 = -3.615-15.6 = -0.616-15.6 = 0.414-15.6 = -1.616-15.6 = 0.416-15.6 = 0.418-15.6 = 2.419-15.6 = 3.4第 3 步：之后我想應(yīng)用每個(gè)結(jié)果的平方1.4 * 1.4 = 1.96-2.6 * -2.6 = 6.76-3.6 * -3.6 = 12.96-0.6 * -0.6 = 0.360.4 * 0.4 = 0.16-1.6 * -1.6 = 2.560.4 * 0.4 = 0.160.4 * 0.4 = 0.162.4 * 2.4 = 5.763.4 * 3.4 = 11.56第 4 步：之后我想對(duì)平方求和1.96 + 6.76 + 12.96 0.36 +0.16 + 2.56 + 0.16 + 0.16 + 5.76 + 11.56 = 42.4我嘗試了這種方法，我能夠達(dá)到第 3 步def sumx(x): for i in x: # result=i-sum(x)/len(x) # result=result*result # #result="{:.2f}".format(result) print("{:.2f}".format(result)) total=0 for i in result: total +=i return (total)sumx(x)錯(cuò)誤信息結(jié)果 i 的錯(cuò)誤消息：TypeError: 'float' object is not iterable所需的輸出為 42.4
3 回答
catspeake
TA貢獻(xiàn)1111條經(jīng)驗(yàn) 獲得超0個(gè)贊
您可以使用numpy并列出理解：
import numpy as np
x=[17,13,12,15,16,14,16,16,18,19]
# using mean function from numpy
sum((y - np.mean(x))**2 for y in x) # 42.4
# calculating mean on our own (pure python)
sum((y - (sum(x)/float(len(x))))**2 for y in x) # 42.4
# function to calculate sse
def sse(x):
m = np.mean(x)
return sum((y - m)**2 for y in x)
繁華開滿天機(jī)
TA貢獻(xiàn)1816條經(jīng)驗(yàn) 獲得超4個(gè)贊
您可以使用numpy和矢量化計(jì)算：
import numpy as np
x = np.array([17,13,12,15,16,14,16,16,18,19])
normalized_vector = x - np.mean(x)
result = np.dot(normalized_vector, normalized_vector)
...因?yàn)辄c(diǎn)積等于元素的平方和。更簡潔：
result = np.var(x) * x.size
...計(jì)算n倍方差，這應(yīng)該等于你想要的。
慕萊塢森
TA貢獻(xiàn)1810條經(jīng)驗(yàn) 獲得超4個(gè)贊
你的錯(cuò)誤在這里：
total=0
for i in result:
total +=i
這里的結(jié)果只是一個(gè)數(shù)字而不是一個(gè)列表。因此整個(gè)代碼都寫錯(cuò)了。
相反，您可以在函數(shù)的起始位置初始化 total 并擺脫 for 循環(huán)。
這是一個(gè)簡單的代碼：
x = [17,13,12,15,16,14,16,16,18,19]
average = sum (x) / float (len (x))
total = 0
for number in x:
result = pow ((number - average), 2)
total += result
print total
如果你想把它作為一個(gè)函數(shù)：
x = [17,13,12,15,16,14,16,16,18,19]
def sumx(x):
average = sum (x) / float (len (x))
total = 0
for i in x:
result = i - average
result = result * result
total += result
return (total)
print sumx(x)
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