【发布时间】:2021-12-07 23:23:58
【问题描述】:
我正在做一个从列表中查找中位数的简短函数。
这是 csv 文件的一部分:
Index,Height(Inches),Weight(Pounds)
1,65.78331,112.9925
2,71.51521,136.4873
3,69.39874,153.0269
4,68.2166,142.3354
5,67.78781,144.2971
6,68.69784,123.3024
7,69.80204,141.4947
8,70.01472,136.4623
9,67.90265,112.3723
10,66.78236,120.6672
11,66.48769,127.4516
12,67.62333,114.143
13,68.30248,125.6107
14,67.11656,122.4618
15,68.27967,116.0866
16,71.0916,139.9975
17,66.461,129.5023
18,68.64927,142.9733
19,71.23033,137.9025
20,67.13118,124.0449
21,67.83379,141.2807
22,68.87881,143.5392
23,63.48115,97.90191
24,68.42187,129.5027
25,67.62804,141.8501
26,67.20864,129.7244
27,70.84235,142.4235
有人可以帮助我吗?
我也尝试使用Counter 来计算项目数。
我想找到第三列的中位数。
我预先存在的功能是:
def median():
n = (len(file_data))
file_data.sort()
if n%2==0:
median1 = file_data[n//2]
median2 = file_data[n//2-1]
median = (median1+median2)/2
mediankg1 = median/2.2046
else:
median = file_data[n//2]
mediankg = median/2.2046
print("MEDIAN")
print("Median is " + str(median)+" pounds")
print("OR")
print("Median is " + str(mediankg1)+" kilograms")
median()
【问题讨论】:
-
用熊猫怎么样?
-
您应该将自己尝试使用
Counter添加到您的问题中。我没有downvote your question because no attempt was made,因为您是新贡献者,但通常我们希望您至少提出honest attempt at the solution,然后然后询问有关您的实施的具体问题。跨度>