python：根据 IF 条件过滤

Question

我正在使用简单的 python 条件进行操作，旨在过滤 > 或等于零的值，并将过滤后的值存储在列表中

# make a object contained all clusters
clustering = d.clusterer.clustering_dict[cut_off]
# list of ignored objects
banned_conf=[]
for clust in clustering:
    clustStr = str(clustering.index(clust))
    clustStr = int(clustStr) + 1
    # get the value of energy for the clust
    ener=clust[0].energy
    # set up filter to ignore conformations with positive energies
    if ener > 0:
        print('Conformation in ' + str(clustStr) + ' cluster poses positive energy')
        banned_conf.append(ener)
        print('Nonsence: It is ignored!')
        continue
    elif ener == 0:
        print('Conformation in ' + str(clustStr) + ' cluster poses ZERO energy')
        banned_conf.append(ener)
        print('Very rare case: it is ignored!')
        continue
    #else:
        #print("Ain't no wrong conformations in "  + str(clustStr) + " cluster")

怎么可能在同一个 IF 语句（没有 elif）中忽略所有 > 或 = 0 的值？ 哪种过滤会更好（使用 elif 或在单个 IF 语句中）？

Answer 1

您可以使用>=同时测试这两个条件。

for index, clust in enumerate(clustering, 1):
    ener = clust[0].energy
    if ener >= 0:
        print(f'Conformation in {index} cluster poses zero or positive energy, it is ignored')
        banned_conf.append(clust)

如果您想为零能量和正能量显示不同的信息，您的原始方法会更好。

Answer 2

我会使用filter function：

lst = [0,1,-1,2,-2,3,-3,4,-4]
filtered = list(filter(lambda x: x >= 0, lst))
for ele in filtered:
    print(f'{ele} is >= 0')

或者，如果您不想使用 lamda function 和过滤器，我会这样做：

lst = [0,1,-1,2,-2,3,-3,4,-4]
filtered = []
for ele in lst:
    if ele >= 0:
        filtered.append(ele)
for ele in filtered:
    print(f'{ele} is >= 0')

或者您可以使用列表理解：

lst = [0,1,-1,2,-2,3,-3,4,-4]
filtered = [for ele in lst if ele >= 0]
for ele in filtered:
    print(f'{ele} is >= 0')