我用ID /里程表/日期创建了排序矩阵。我真的需要删除所有列表，其中第二个位置为0值。 我准备了一个函数和循环，但不幸的是我仍然低于错误。我怀疑del []方法是否正确使用？
zero_clear中，如果tablica [i] [1] == 0：****
tablica = [
[
[21091705, 0,'30.06.2019'],
[21091705, 0,'29.06.2019'],
[21091705, 653842,'29.06.2019'],
[21091705, 641701,'28.06.2019'],
[21091705, 642000,'27.06.2019'],
[21091705, 0,'25.06.2019'],
[21091705, 642030,'25.06.2019'],
[21091705, 642030,'25.06.2019'],
[21091705, 637030,'25.06.2019'],
[21091705, 0,'25.06.2019'],
[21091705, 0,'17.05.2019'],
[21091705, 641256,'25.06.2019']
],
[
[21091703, 487000,'30.06.2019'],
[21091703, 461524,'29.06.2019'],
[21091703, 0,'29.06.2019'],
[21091703, 0,'28.06.2019'],
[21091703, 427701,'28.06.2019'],
[21091703, 429000,'27.06.2019']
],
[
[21091701, 421077,'30.06.2019'],
[21091701, 453777,'29.06.2019'],
[21091701, 0,'29.06.2019'],
[21091701, 427701,'28.06.2019'],
[21091701, 433123,'27.06.2019']
],
[
[21091704, 491000,'30.06.2019'],
[21091704, 0,'29.06.2019'],
[21091704, 474542,'29.06.2019'],
[21091704, 427701,'28.06.2019'],
[21091704, 427701,'28.06.2019'],
[21091704, 0,'27.06.2019']
]
]
def zero_clear(tablica):
for i in range(len(tablica)):
if tablica[i][1] == 0:
del tablica[i]
return tablica
for item in tablica:
zero_clear(item)
答案 0 :(得分：1)
您在迭代时减小了列表大小，因此遇到了这个问题。尝试下面的代码并检查！
const std::vector<int> myVector = {1, 2, 3};
const std::vector<int> myVector2{1, 2, 3};
答案 1 :(得分：1)
如果在迭代（使用tablica的过程中从for列表元素中删除元素，您将结束为每个删除的元素跳转下一个元素，以解决此问题，您可以使用while循环或创建新列表：
1）使用while循环
def zero_clear(tablica):
i = 0
while i < len(tablica):
if tablica[i][1] == 0:
del tablica[i]
else:
i += 1
for item in tablica:
zero_clear(item)
2）创建一个新列表
def zero_clear(tablica):
new_list = []
for l in tablica:
if l[1] != 0:
new_list.append(l)
return new_list
for item in tablica:
item[:] = zero_clear(item)
或通过使用列表理解来创建新元素：
for item in tablica:
item[:] = [l for l in item if l[1] != 0]
或使用列表理解功能创建一个全新的列表：
tablica = [[l for l in item if l[1] != 0] for item in tablica]
输出：
[[[21091705, 653842, '29.06.2019'],
[21091705, 641701, '28.06.2019'],
[21091705, 642000, '27.06.2019'],
[21091705, 642030, '25.06.2019'],
[21091705, 642030, '25.06.2019'],
[21091705, 637030, '25.06.2019'],
[21091705, 641256, '25.06.2019']],
[[21091703, 487000, '30.06.2019'],
[21091703, 461524, '29.06.2019'],
[21091703, 427701, '28.06.2019'],
[21091703, 429000, '27.06.2019']],
[[21091701, 421077, '30.06.2019'],
[21091701, 453777, '29.06.2019'],
[21091701, 427701, '28.06.2019'],
[21091701, 433123, '27.06.2019']],
[[21091704, 491000, '30.06.2019'],
[21091704, 474542, '29.06.2019'],
[21091704, 427701, '28.06.2019'],
[21091704, 427701, '28.06.2019']]]
答案 2 :(得分：0)
您的问题是您在阅读表格时正在删除项目。每次删除项目时，表格都会变短，因此最终会出现list index out of range错误。
您可以改用这种方式：
tablica = [[[row for i in range(len(item))] for row in item if row[1] != 0] for item in tablica]
答案 3 :(得分：0)
这是使用Pods库的解决方案：
from fyers_api import accessToken
from fyers_api import fyersModel
app_id = "xxxxxxxxxx"
app_secret = "xxxxxxxxxx"
app_session = accessToken.SessionModel(app_id, app_secret)
response = app_session.auth()
print(app_session)
print(response)
authorization_code = response['data']['authorization_code']
app_session.set_token(authorization_code)
gen_token = app_session.generate_token()
print("token url is copy paste this url in browser and copy access
token excluding your id at Last ")
print(gen_token)
print("tokent printed thanks")
token="gAAAAABeTWk7AnufuuQQx0D0NkgABinWk7AnufuuQQx0DQ3ctAFWk7AnufuuQQx0DMQQwacJ-
_xUVnrTu2Pk5K5QCLF0SZmw7nlpaWk7AnufuuQQx0DG4_3EGCYw92-iAh8="
is_async = False
fyers = fyersModel.FyersModel(is_async)
print(fyers. get_profile(token = token))
fyers.funds(token = token)
print(fyers.funds(token = token))
我还没有测试时序，但是通过这种方式，该解决方案可以轻松扩展到大型矩阵。我建议您使用pandas库来管理类似表格的数据，因为它具有许多比纯Python更有效的方法。
答案 4 :(得分：0)
我觉得出了点问题，因为在更改矩阵后，当我使用简单的for循环时，要看到输出，应该是这样的（第二个位置没有列表，且位置为0）
for ele in tablica:
for item in ele:
print(item)
[21091705, 653842, '29.06.2019']
[21091705, 641701, '28.06.2019']
[21091705, 642000, '27.06.2019']
[21091705, 642030, '25.06.2019']
[21091705, 642030, '25.06.2019']
[21091705, 637030, '25.06.2019']
[21091705, 641256, '25.06.2019']
[21091703, 487000, '30.06.2019']
[21091703, 461524, '29.06.2019']
[21091703, 427701, '28.06.2019']
[21091703, 429000, '27.06.2019']
[21091701, 421077, '30.06.2019']
[21091701, 453777, '29.06.2019']
[21091701, 427701, '28.06.2019']
[21091701, 433123, '27.06.2019']
[21091704, 491000, '30.06.2019']
[21091704, 474542, '29.06.2019']
[21091704, 427701, '28.06.2019']
[21091704, 427701, '28.06.2019']
当我使用此循环查看矩阵更改后的输出时，输出要复杂得多：
tablica = [[[row for i in range(len(item))] for row in item if row[1] != 0] for item in tablica]
输出：
for ele in tablica:
for item in ele:
print(item)
[[21091705, 653842, '29.06.2019'], [21091705, 653842, '29.06.2019'], [21091705, 653842, '29.06.2019'], [21091705, 653842, '29.06.2019'], [21091705, 653842, '29.06.2019'], [21091705, 653842, '29.06.2019'], [21091705, 653842, '29.06.2019'], [21091705, 653842, '29.06.2019'], [21091705, 653842, '29.06.2019'], [21091705, 653842, '29.06.2019'], [21091705, 653842, '29.06.2019'], [21091705, 653842, '29.06.2019']]
[[21091705, 641701, '28.06.2019'], [21091705, 641701, '28.06.2019'], [21091705, 641701, '28.06.2019'], [21091705, 641701, '28.06.2019'], [21091705, 641701, '28.06.2019'], [21091705, 641701, '28.06.2019'], [21091705, 641701, '28.06.2019'], [21091705, 641701, '28.06.2019'], [21091705, 641701, '28.06.2019'], [21091705, 641701, '28.06.2019'], [21091705, 641701, '28.06.2019'], [21091705, 641701, '28.06.2019']]
[[21091705, 642000, '27.06.2019'], [21091705, 642000, '27.06.2019'], [21091705, 642000, '27.06.2019'], [21091705, 642000, '27.06.2019'], [21091705, 642000, '27.06.2019'], [21091705, 642000, '27.06.2019'], [21091705, 642000, '27.06.2019'], [21091705, 642000, '27.06.2019'], [21091705, 642000, '27.06.2019'], [21091705, 642000, '27.06.2019'], [21091705, 642000, '27.06.2019'], [21091705, 642000, '27.06.2019']]