如果和elif重复

问题描述 投票:1回答:2

我想知道是否有办法使这种类型的代码更简洁和自动化,这样我就不必一次又一次地重复这些elif语句。

ga是从1记录到我记录的任何秒数的累积秒数列表,我想用每小时与相应的数据帧分开那个小时有没有办法让这更简单?

for i in range(len(ga)):
 if ga[i]<=3600:
    j.append(i)
 elif ga[i]<=7200:
    u.append(i)
 elif ga[i]<=10800:
    k.append(i)
 elif ga[i]<=14400:
    b.append(i)
 elif ga[i]<=18000:
    bi.append(i)
 elif ga[i]<=21600:
    bit.append(i)
 elif ga[i]<=25200:
    bitb.append(i)
 elif ga[i]<=28800:
    bitc.append(i)
 elif ga[i]<=32400:
    bitd.append(i)
 elif ga[i]<=36000:
    bite.append(i)
python-3.x computer-science custom-lists
2个回答
0
投票

使用列表推导:

r = [[idx for idx, val in enumerate(
    ga) if 3600*level < val <= 3600*(1+level)] for level in range(9)]

(j, u, k, b, bi, bit, bitc, bitd, bite) = r

1
投票

也许,它不是一般的解决方案,但问题的基础,你可以做出创造性的解决方案。像下面的代码

# Put variable name in var
# These are empty lists that have already been defined
var = ["j", "u", "k", "b", "bi", "bit", "bitb", "bitc", "bitd", "bite"]
for i in range(len(ga)):
    # Finding var index base of ga[i] value.
    var_index = ga[i] // 3600
    if ga[i] % 3600 == 0 and ga[i] > 0:
        var_index -= 1
    # Add the ga index to the corresponding list
    eval(var[var_index] + ".append(" + str(i) + ")")

但最好使用字典而不是定义许多变量。像下面的代码。

# Define dictionary 
ch = {"j":[], "u":[], "k":[], "b":[], "bi":[],
      "bit":[], "bitb":[], "bitc":[], "bitd":[], "bite":[]}
# Dictionary keys
key = ["j", "u", "k", "b", "bi", "bit", "bitb", "bitc", "bitd", "bite"]
for i in range(len(ga)):
    # Finding key index base of ga[i] value.
    key_index = ga[i] // 3600
    if ga[i] % 3600 == 0 and ga[i] > 0:
        key_index -= 1
    # Add the ga index to the corresponding list in the dictionary
    ch[key[key_index]].append(i)
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