Pandas - 检查数据框列是不是包含字典中的键：值对

Posted 2023-03-11

【中文标题】Pandas - 检查数据框列是不是包含字典中的键：值对

【英文标题】：Pandas - check if dataframe columns contain key:value pairs from a dictionaryPandas - 检查数据框列是否包含字典中的键：值对

【发布时间】：2017-09-12 16:42:00

【问题描述】：

这个问题与我发布的另一个问题有关。 Pandas - check if a string column in one dataframe contains a pair of strings from another dataframe

我的目标是检查数据框的两个不同列是否包含一对字符串值，如果满足条件，则提取其中一个值。

我有两个这样的数据框：

df1 = pd.DataFrame('consumption':['squirrelate apple', 'monkey likesapple', 
                                  'monkey banana gets', 'badger/getsbanana', 'giraffe eats grass', 'badger apple.loves', 'elephant is huge', 'elephant/eats/', 'squirrel.digsingrass'], 
                    'name': ['apple', 'appleisred', 'banana is tropical', 'banana is soft', 'lemon is sour', 'washington apples', 'kiwi', 'bananas', 'apples'])

df2 = pd.DataFrame('food':['apple', 'apple', 'banana', 'banana'], 'creature':['squirrel', 'badger', 'monkey', 'elephant'])

In [187]:df1
Out[187]: 
            consumption                name
0     squirrelate apple               apple
1     monkey likesapple          appleisred
2    monkey banana gets  banana is tropical
3     badger/getsbanana      banana is soft
4    giraffe eats grass       lemon is sour
5    badger apple.loves   washington apples
6      elephant is huge                kiwi
7        elephant/eats/             bananas
8  squirrel.digsingrass              apples

In[188]: df2
Out[188]: 
   creature    food
0  squirrel   apple
1    badger   apple
2    monkey  banana
3  elephant  banana

我想要做的是测试'apple'是否出现在df1['name']中和'squirrel'出现在df1['consumption']中，如果这两个条件都满足然后从df1['consumption']中提取'squirrel'到一个新列df['creature']中.结果应如下所示：

Out[189]: 
            consumption  creature                name
0     squirrelate apple  squirrel               apple
1     monkey likesapple       NaN          appleisred
2    monkey banana gets    monkey  banana is tropical
3     badger/getsbanana       NaN      banana is soft
4    giraffe eats grass       NaN       lemon is sour
5    badger apple.loves    badger   washington apples
6      elephant is huge       NaN                kiwi
7        elephant/eats/  elephant             bananas
8  squirrel.digsingrass       NaN              apples

如果没有配对值约束，我可以做一些简单的事情，比如：

np.where((df1['consumption'].str.contains(<creature_string>, case = False)) & (df1['name'].str.contains(<food_string>, case = False)), df['consumption'].str.extract(<creature_string>), np.nan)

但我必须检查对所以我尝试将食物作为键和生物作为值的字典，然后为给定的食物键创建所有生物的字符串 var 并查找使用 str.contains 的那些：

unique_food = df2.food.unique()
food_dict = elem : pd.DataFrame for elem in unique_food
for key in food_dict.keys():
    food_dict[key] = df2[:][df2.food == key]

# create key:value pairs of food key and creature strings
food_strings = 
for key, values in food_dict.items():
    food_strings.update(key: '|'.join(map(str, list(food_dict[key]['creature'].unique()))))

In[199]: food_strings
Out[199]: 'apple': 'squirrel|badger', 'banana': 'monkey|elephant'

问题是当我现在尝试应用 str.contains 时：

for key, value in food_strings.items():
    np.where((df1['name'].str.contains('('+food_strings[key]+')', case = False)) & 
             (df1['consumption'].str.contains('('+food_strings[value]+')', case = False)), df1['consumptions'].str.extract('('+food_strings[value]+')'), np.nan)

我得到一个 KeyError: 。

---------------------------------------------------------------------------
KeyError                                  Traceback (most recent call last)
<ipython-input-62-7ab718066040> in <module>()
      1 for key, value in food_strings.items():
      2     np.where((df1['name'].str.contains('('+food_strings[key]+')', case = False)) & 
----> 3              (df1['consumption'].str.contains('('+food_strings[value]+')', case = False)), df1['consumption'].str.extract('('+food_strings[value]+')'), np.nan)

KeyError: 'squirrel|badger'

当我只尝试值而不是键时，它适用于第一个键：值对但不适用于第二个：

for key in food_strings.keys():
    df1['test'] = np.where(df1['consumption'].str.contains('('+food_strings[key]+')', case =False), 
                                df1['consumption'].str.extract('('+food_strings[key]+')', expand=False), 
                                np.nan)

df1
Out[196]: 
            consumption                name      test
0     squirrelate apple               apple  squirrel
1     monkey likesapple          appleisred       NaN
2    monkey banana gets  banana is tropical       NaN
3     badger/getsbanana      banana is soft    badger
4    giraffe eats grass       lemon is sour       NaN
5    badger apple.loves   washington apples    badger
6      elephant is huge                kiwi       NaN
7        elephant/eats/             bananas       NaN
8  squirrel.digsingrass              apples  squirrel

我得到了匹配 apple 和 squirrel|badger 的，但错过了 banana:monkey|elephant。

有人可以帮忙吗？

【问题讨论】：

我认为，food_dict 的每个值都包含数据帧而不是字符串。循环进入for key, value in food_dict.items(): 时会发生错误。您将value 作为数据框提供给food_strings[value]。

@titipat 这是错别字对不起-但很好。我编辑了问题并粘贴了我得到的确切错误。

【参考方案1】：

d1 = df1.dropna()
d2 = df2.dropna()

sump = d1.consumption.values.tolist()
name = d1.name.values.tolist()
cret = d2.creature.values.tolist() 
food = d2.food.values.tolist()

check = np.array(
    [
        [c in s and f in n for c, f in zip(cret, food)]
        for s, n in zip(sump, name)
    ]
)

# create a new series with the index of `d1` where we dropped na
# then reindex with `df1.index` prior to `assign`
test = pd.Series(check.dot(d2[['creature']].values).ravel(), d1.index)
test = test.reindex(df1.index, fill_value='')
df1.assign(test=test)

            consumption                name      test
0     squirrelate apple               apple  squirrel
1     monkey likesapple          appleisred          
2    monkey banana gets  banana is tropical    monkey
3     badger/getsbanana      banana is soft          
4    giraffe eats grass       lemon is sour          
5    badger apple.loves   washington apples    badger
6      elephant is huge                kiwi          
7        elephant/eats/             bananas  elephant
8  squirrel.digsingrass              apples  squirrel

【讨论】：

嗨！谢谢 - 很棒的解决方案。一个问题 - 当列表包含 None 值时，它会中断。我收到此错误：TypeError: argument of type 'NoneType' is not iterable。我列出了一个没有 Nonetypes 的列表，sump = df1[df1.consumption.notnull()]['consumption'].values.tolist() 用于油底壳、姓名、cret 和食物。然后 check 函数起作用，但在 df1.assign 中，我得到：ValueError: Length of values does not match length of index

在遍历 zip(sump, name) 时，当 c - s 为非类型或 f - n 均为非类型时，我必须以某种方式获得 NaN / None 值。

dropna 不起作用。 . .然后我会改变数据框！

作为更新：当我这样做时：test = test.reindex(df1.index, fill_value='')df1.assign(test=test)，重新索引会创建多个匹配项 - 例如：squirrelsquirrelsquirrelsquirrel。我这样做了：test = pd.DataFrame(check.dot(d2[['creature']].values).ravel(), d1.index, columns=['some_var'])，然后执行了索引合并：d1 = d1.merge(test, how='left', left_index = True, right_index = True) - 这有意义吗？

合并是有道理的...但是，join 默认情况下在索引上“合并”并且可能会更好....但是/但是，您不应该在重新索引上获得多个匹配项除非您首先在 df1 中有一个非唯一索引。由于您已经展示了您的数据，但您没有展示。