python -如何为字典创建一个更紧凑的组

Question

嗨，我的生物项目代码的这一部分：

# choosing and loading the file:
df = pd.read_csv('Dafniyot_Data.csv',delimiter=',')
#grouping data by C/I groups:
CII = df[df['group'].str.contains('CII')]
CCI = df[df['group'].str.contains('CCI')]
CCC = df[df['group'].str.contains('CCC')]
III = df[df['group'].str.contains('III')]
CIC = df[df['group'].str.contains('CIC')]
ICC = df[df['group'].str.contains('ICC')]
IIC = df[df['group'].str.contains('IIC')]
ICI = df[df['group'].str.contains('ICI')]
#creating a dictonary of the groups:
dict = {'CII':CII, 'CCI':CCI, 'CCC':CCC,'III':III,'CIC':CIC,'ICC':ICC,'IIC':IIC,'ICI':ICI}

#T test
#FERTUNITY
#using ttest for checking FERTUNITY - grandmaternal(F0)
t_F0a = stats.ttest_ind(CCC['N_offspring'],ICC['N_offspring'],nan_policy='omit')
t_F0b = stats.ttest_ind(CCI['N_offspring'],ICI['N_offspring'],nan_policy='omit')
t_F0c = stats.ttest_ind(IIC['N_offspring'],CIC['N_offspring'],nan_policy='omit')
t_F0d = stats.ttest_ind(CCI['N_offspring'],III['N_offspring'],nan_policy='omit')
t_F0 = {'FERTUNITY - grandmaternal(F0)':[t_F0a,t_F0b,t_F0c,t_F0d]}

我需要多次重复ttest第6部分，要么更改组(CCC等)，要么更改df(‘N_后代’，生存)中的行，这需要在项目中使用很多行。

我正试图找到一种方法，最终仍能得到每一组的字典：

t_F0 = {'FERTUNITY - grandmaternal(F0)':[t_F0a,t_F0b,t_F0c,t_F0d]}

因为它对我以后有用，但以一种较少重复的方式，用较少的线条

Answer 1

使用itertools.product生成所有键，使用dict理解生成值：

from itertools import product
keys = [''.join(items) for items in product("CI", repeat=3)]
the_dict = { key: df[df['group'].str.contains(key)] for key in keys }

类似地，您可以生成测试键的后半部分：

half_keys = [''.join(items) for items in product("CI", repeat=2)]
t_F0 = {
    'FERTUNITY - grandmaternal(F0)': [
        stats.ttest_ind(
            the_dict[f"C{half_key}"]['N_offspring'],
            the_dict[f"I{half_key}"]['N_offspring'],
            nan_policy='omit'
        ) for half_key in half_keys
    ],
}

顺便说一句，您不应该使用dict作为变量名:它已经有了意义( dict对象的类型)。

暂且不说，这涉及如何干涸创建词典这一字面问题。但是，请考虑克里斯在评论中所说的话；这可能是一个XY problem。