将数据框列与二维列表 python 进行比较答案

【问题标题】：compare a dataframe column with 2D list python将数据框列与二维列表 python 进行比较
【发布时间】：2020-08-14 12:24:25
【问题描述】：

我有一个数据框data：

Stretch
SB
NB
...
WB

行将有Nan。我有 2D 列表：

[['SB', someFloatvalue, someFloatvalue], ['SB', someFloatvalue, someFloatvalue], ['NB', someFloatvalue, someFloatvalue]]

[['NB', someFloatvalue, someFloatvalue], ['SB', someFloatvalue, someFloatvalue], ['NB', someFloatvalue, someFloatvalue]]

上面的每个二维列表都与Stretch 中的每一行相关联。我想删除所有与Strecth 值不同的列表。最后，预期的结果是：

[['SB', someFloatvalue, someFloatvalue], ['SB', someFloatvalue, someFloatvalue]]

[['NB', someFloatvalue, someFloatvalue], ['NB', someFloatvalue, someFloatvalue]]

说明： Stretch 的第一行是 SB。因此，相应的 2D 列表应该只列出 SB。等等... 在Stretch 中的Nan 的情况下，只需按原样显示列表即可。

我的想法是将Stretch 转换为tuple，因为我认为这个位置很重要。它没有成功..我不知道如何映射它们..

【问题讨论】：

标签： python pandas list dataframe tuples

【解决方案1】：

您可以尝试在list-comprehension中使用zip()：

out = [[l for l in sub_data if l[0] == key or str(key) == "nan" ] for sub_data, key in zip(data, df.Stretch)]

或者在通常的for循环中：

out_1 = []
for sub_data, key in zip(data, df.Stretch):
    sub = []
    for l in sub_data:
        if l[0] == key or str(key) == "nan":
            sub.append(l)
    if len(sub) > 0: out_1.append(sub)

完整代码

df = pd.DataFrame({"Stretch": ["SB", "NB", np.NaN]})

data = [[['SB', 123456, 123456], ['SB', 123456, 123456], ['NB', 123456, 123456]],
        [['NB', 123456, 123456], ['SB', 123456, 123456], ['NB', 123456, 123456]],
        [['NB', 123456, 123456], ['SB', 123456, 123456], ['NB', 123456, 123456]]]

# Usual for loop
out_1 = [[l for l in sub_data if l[0] == key or str(key) == "nan"] for sub_data, key in zip(data, df.Stretch)]
print(out_1)
# [[['SB', 123456, 123456], ['SB', 123456, 123456]],
#  [['NB', 123456, 123456], ['NB', 123456, 123456]],
#  [['NB', 123456, 123456], ['SB', 123456, 123456], ['NB', 123456, 123456]]]


out_2 = []
for sub_data, key in zip(data, df.Stretch):
    sub = []
    for l in sub_data:
        if l[0] == key or str(key) == "nan":
            sub.append(l)
    if len(sub) > 0: out_2.append(sub)

print(out_2)
# [[['SB', 123456, 123456], ['SB', 123456, 123456]],
#  [['NB', 123456, 123456], ['NB', 123456, 123456]],
#  [['NB', 123456, 123456], ['SB', 123456, 123456], ['NB', 123456, 123456]]]

print(out_1 == out_2)
# True

【讨论】：