需求分析
根据原始数据，计算出累计和、回撤、连续正确、连续错误、连续正确值与连续错误值6项数据，其中原始数据大于等于0认定为正确，原始数据小于0为错误。明白了要求，那我们就开始撸代码吧~
解决步骤
import pandas as pd
#创建一个计算数据的函数
def calculate(df):
pass
#读取原始数据，将索引列去除
df = pd.read_excel('需求0621.xlsx',index_col=0)
#调用计算数据的函数
calculate(df)先把整体思路写好，再去想办法计算每项数据
#计算累计和
lst1 = []
sum = 0
for i in range(df.shape[0]):
if i == 0:
lst1.append(df['N'][i])
sum = df['N'][i]
else:
sum = df['N'][i]
lst1.append(sum)
df['累计和'] = lst1#计算回撤
lst2 = []
max = 0
for i in range(df.shape[0]):
if i == 0:
lst2.append(0)
elif df['累计和'][i] > max:
max = df['累计和'][i]
lst2.append(0)
elif df['累计和'][i] < max:
lst2.append(df['累计和'][i]-max)
elif df['累计和'][i] == max:
lst2.append(0)
df['回撤'] = lst2#计算连续正确的个数
lst3 = []
correct = 0
for i in range(df.shape[0]):
if df['N'][i] >= 0:
correct = 1
lst3.append(correct)
else:
lst3.append(0)
correct = 0
df['连续正确'] = lst3#计算连续错误的个数
lst4 = []
mistake = 0
for i in range(df.shape[0]):
if df['N'][i] < 0:
mistake = 1
lst4.append(mistake)
else:
lst4.append(0)
mistake = 0
df['连续错误'] = lst4#计算连续正确值
lst5 = []
for i in range(df.shape[0]):
lst5.append('')
right = 0
for i in range(df.shape[0]):
if df['连续正确'][i] != 0:
right = df['N'][i]
elif df['连续正确'][i] == 0 and right != 0:
lst5[i-1] = right
right = 0
df['连续正确值'] = lst5#计算连续错误值
lst6 = []
for i in range(df.shape[0]):
lst6.append('')
wrong = 0
for i in range(df.shape[0]):
if df['连续错误'][i] != 0:
wrong = df['N'][i]
elif df['连续错误'][i] == 0 and wrong != 0:
lst6[i-1] = wrong
wrong = 0
df['连续错误值'] = lst6最后将dataframe保存到excel
df.to_excel('完成计算.xlsx')
print('保持成功')最终结果
技术总结
虽然已经完成了要求计算出了所有的数据，但在写代码过程中计算的步骤都是基于python基础语法实现的，对于pandas的使用还要只有通过大量的练习才能够熟练的掌握
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