如何解决如何从方法中获取赋值？
我想在设置某个字段的类中收集方法，例如
import numpy as np
import itertools
from collections import Counter
import pandas as pd
bag = np.hstack((
np.repeat(0,80),np.repeat(1,21),np.repeat(3,5),np.repeat(7,1)
))
#107*106*105*104*103*102*101*100*99*98
#Out[176]: 127506499163211168000 #Permutations
##Need to reduce the number to sample from without changing the number of possible combinations
reduced_bag = np.hstack((
np.repeat(0,10),## 0 can be chosen all 10 times
np.repeat(1,## 1 can be chosen all 10 times
np.repeat(3,## 3 can be chosen up to 5 times
np.repeat(7,1) ## 7 can be chosen once
))
## There are 96 unique combinations
number_unique_combinations = len(set(list(itertools.combinations(reduced_bag,10))))
### sorted list of all combinations
unique_combinations = sorted(list(set(list(itertools.combinations(reduced_bag,10)))))
### sum of each unique combination
sums_list = [sum(uc) for uc in unique_combinations]
### probability for each unique combination
probability_dict = {0:80,1:21,3:5,7:1} ##Dictionary to refer to
n = 107 ##Number in the bag
probability_list = []
##This part is VERY slow to run because of the itertools.permutations
for x in unique_combinations:
print(x)
p = 1 ##Start with the probability again
n = 107 ##Start with a full bag for each combination
count_x = Counter(x)
for i in x:
i_left = probability_dict[i] - (Counter(x)[i] - count_x[i]) ##Number of that type left in bag
p *= i_left/n ##Multiply the probability
n = n-1 # non replacement
count_x[i] = count_x[i] - 1 ##Reduce the number in the bag
p *= len(set(list(itertools.permutations(x,10)))) ##Multiply by the number of permutations per combination
probability_list.append(p)
##sum(probability_list) ## Has a rounding error
##Out[57]: 1.0000000000000002
##
##Put the combinations into dataframe
ar = np.array((unique_combinations,sums_list,probability_list))
df = pd.DataFrame(ar).T
##Name the columns
df.columns = ["combination","sum","probability"]
## probability that sum is >= 6
df[df["sum"] >= 6]['probability'].sum()
## 0.24139909236232826
## probability that sum is == 6
df[df["sum"] == 6]['probability'].sum()
## 0.06756408790812335
。
我知道如何获取方法的 int i = 2;：
Statement但 assignments 似乎是表达式，我不知道如何找到它们。如何在方法中获取（赋值）表达式？
解决方法
if (statemet.isExpressionStmt()) {
Expression e = statement.asExpressionStmt().getExpression();
成功了。
版权声明：本文内容由互联网用户自发贡献，该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务，不拥有所有权，不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容， 请发送邮件至 dio@foxmail.com 举报，一经查实，本站将立刻删除。
