从列表中选择子列表，其中子列表的项与给定值的差异最小

Question

I'm using python v2.7 and I have the parameter mean_val=35.3085666667 I have a list new_add_list that contains sublists. In new_add_list[3] I have a score value. From new_add_list I want to get the sublist that has the minimun value of abs(new_add_list[3]-mean_val) .

new_add_list=[
['encoding_txt\\pirkagia_10b_192x144_15_qp_26_ON_encoder_randomaccess_B16_Filte
rs_ON_scalable_JOURNAL2017.txt', '192x144', 26, 35.1782, 136.8125, 300.788, 0.844, -3.5], 
['encoding_txt\\pirkagia_10b_384x288_15_qp_28_ON_encoder_randomaccess_
B16_Filters_ON_scalable_JOURNAL2017.txt', '384x288', 28, 35.2193, 275.4614, 1065.412, 2.167, -2], 
['encoding_txt\\pirkagia_10b_720x576_15_qp_30_ON_encoder_rando
maccess_B16_Filters_ON_scalable_JOURNAL2017.txt', '720x576', 30, 35.5282, 484.3387, 3536.937, 6.679, 1.5]]

For that example the answer must be:

['encoding_txt\\pirkagia_10b_384x288_15_qp_28_ON_encoder_randomaccess_
B16_Filters_ON_scalable_JOURNAL2017.txt', '384x288', 28, 35.2193, 275.4614, 1065.412, 2.167, -2]

Can I use something like sub(new_add_list, key=itemgetter(3)-mean_val)

Answer 1

将min与适当的按键功能配合使用：

min(new_add_list, key=lambda l: abs(l[3] - mean_val))