Python读取txt文件并操作时，其文件名中有小数，总是报错，如何解决？

我的文件名中有1.000 1.050 1.100这样一直到2，我希望按顺序打开并进行操作，但是总是报错float object cannot be interpreted as an integer。我该怎么改第三行代码可以让它实现open

1. 

2. 代码：
   import os
   for i in range(1,2,0.05):
   with open(f'D:/test/test/a{＂%.3f＂i}', 'r') as f:

    context = f.readlines()
   

   target_line = 0
   for l in context:

    if l.find('test') != -1:
        break
    target_line += 1
   

   with open('D:/test/out.txt', 'a') as file1:

    print( context[target_line + 15], file=file1)
    print( context[target_line + 16], file=file1)
   

range不支持小数吧，自己组合下就行

import os
lst=['a','b']#更多文件前缀
for c in lst:
    for i in range(1000,2001,50):
        file='D:/test/test/'+c+str(i)[0]+"."+str(i)[1:]
        print(file)
        #with open(file) as f:
        #....其他代码逻辑

range()不支持小数,用while 循环即可

import os
li = ['a','b'] #文件前面字符串
for s in li:
    i = 1
    while i<=2.00001:
        file = f'D:/test/test/{s}{i:.3f}'
        print(file)
        # with open(file, 'r') as f:
        #     context = f.readlines()
        i+=0.05

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for 后面放
a = a + str(i)
然后里面改成{a}就好了