python - 第一次尝试 python，错误（“IndexError: index 8 is out of bounds for axis 0 with size 8”）和效率问题

Question

上周才开始学习 python，已经有 20 年没有用其他方式编码了，而且从一开始就没有那么先进。我把你好世界的事记下来了。现在我试图回测外汇对。感谢您对学习曲线的任何帮助，当然还有在我的 Lynda 视频上搜索这个网站。

遇到一个奇怪的错误，并且想知道是否有更有效的方法可以像我一样循环遍历 excel 数据列。

正在阅读的电子表格很简单...... A 列下方有 56 个外汇对，列标题为日期的位置上方有 8 行，每列中的单元格是该日期各自的外汇对收盘价。该策略从第 2 列的顶部开始（因此可以计算出相对于先前先验的回报百分比）并计算出每对的周期/周期回报百分比，确定哪个是“最大值”，然后“做多”那个表现最好的人......其在随后的时期/时期的表现被记录为投资组合的 PnL（代码中的“p”），循环直到读取当前最近的列。

该错误与使用 8 列而不是 7 列有关...当我将循环限制为 7 列而不是 8 列时有效。当我使用 8 时，我得到一堵以“IndexError：索引 8 超出轴 0 的范围”结尾的文本墙大小为 8" 当我使用太多行时，类似的错误是 56 而不是 55，我认为我错过了最后一行。

这是我的代码：,,,

enter code here
#set up imports
import pandas as pd 

#import spreadsheet 
x1 = pd.ExcelFile(r"C:\Users\Gamblor\Desktop\Python\test2020.xlsx")
df = pd.read_excel(x1, "Sheet1", header=1)

#define counters for loops
o = 1 # observation counter
c = 3 # column counter
r = 0 # active row counter for sorting through for max

#define identifiers for the portfolio
rpos = 0 # static row, for identifying which currency pair is in column 0 of that row
p = 100 # portfolio size starts at $100

#define the stuff we are evaluating for
pair = df.iat[r,0] # starting pair at 0,0 where each loop will begin
pair_pct_rtn = 0 # starts out at zero, becomes something at first evaluation, then gets 
compared to each subsequent eval
pair_pct_rtn_calc = 0 # a second version of above, for comparison to prior return

#runs a loop starting at the top to find the max period/period % return in a specific column

while (c < 8): # manually limiting this to 5 columns left to right

    while (r < 55): # i am manually limiting this to 55 data rows per the spreadsheet ... would be better if automatic

        pair_pct_rtn_calc = ((df.iat[r,c])/(df.iat[r,c-1]) - 1)

        if pair_pct_rtn_calc > pair_pct_rtn: # if its a higher return, it must be the "max" to that point
            pair = df.iat[r,0] # identifies the max pair for this column observation, so far
            pair_pct_rtn = pair_pct_rtn_calc # sets pair_pct_rtn as the new max
            rpos = r # identifies the max pair's ROW for this column observation, so far

        r = r + 1 # adds to r in order to jump down and calc the next row

    print('in obs #', o ,', ', pair ,'did best at' ,pair_pct_rtn ,'.')
    o = o + 1

    # now adjust the portfolio by however well USDMXN did in the subsequent week

    p = p * ( 1 + ((df.iat[rpos,c+1])/(df.iat[rpos,c]) - 1))
    print('then the subsequent period it did: ',(df.iat[rpos,c+1])/(df.iat[rpos,c]) - 1)
    print('resulting in portfolio value of', p)
    rpos = 0
    r = 0
    pair_pct_rtn = 0
    c = c + 1 # adds to c in order to move to the next period to the right

print(p)

标签： pythonloops

Answer 1

由于索引从 0 开始标记，因此您要查找的第 8 个元素的索引为 7。同样，行索引 55（第 56 行）将是您的最后一行。