检查列表中的项目是否在列表类型的列中可用答案

【问题标题】：check if an item in a list is available in a column which is of type list检查列表中的项目是否在列表类型的列中可用
【发布时间】：2020-05-18 19:40:57
【问题描述】：

我正在尝试使用数据框中的一列迭代我拥有的列表，该列在每一行中都有列表。

list1 = ['正在安装','安装','已安装','替换','修复','修复','替换','零件','使用','新']

df[lwr_nopunc_spc_nostpwrd].head(3)

['daily', 'ask', 'questions']
['daily', 'system', 'check', 'task',  'replace']
['inspection', 'complete', 'replaced', 'horizontal', 'sealing', 'blade', 'inspection', 'complete', 'issues', 'found']

现在，我想在我的数据框中获取两个新列，如果 <new column one> df[lwr_nopunc_spc_nostpwrd] 行中的任何一项存在于 list1 <new columns two> 中，如果 list1 中的所有项目为一个，则它们应该显示 true 或 false存在于 df[lwr_nopunc_spc_nostpwrd] 行中

请告诉我如何实现。我尝试了all() 和any() 方法，但似乎不起作用。

def prt_usd(row):
    return(any(item in query['lwr_nopunc_spc_nostpwrd'] for item in part))

for row in query['lwr_nopunc_spc_nostpwrd']:
    prt_usd(query['lwr_nopunc_spc_nostpwrd'])

【问题讨论】：

标签： python-3.x pandas dataframe

【解决方案1】：

你可以做到 apply 和 set 喜欢：

# I changed the list to the second row to show that the column all works
list1 = ['daily', 'system', 'check', 'task',  'replace']
# create a set from it
s1 = set(list1)

# for any word, check that the intersection of s1 
# and the set of the list in this row is not empty
df['col_any'] = df['lwr_nopunc_spc_nostpwrd'].apply(lambda x: any(set(x)&s1))

# for all, subtract the set of this row from the set s1, 
# if not empty then it return True with any
# that you reverse using ~ in front of it to get True if all words from s1 are in this row
df['col_all'] = ~df['lwr_nopunc_spc_nostpwrd'].apply(lambda x: any(s1-set(x)))
print (df)
                             lwr_nopunc_spc_nostpwrd  col_any  col_all
0                            [daily, ask, questions]     True    False
1              [daily, system, check, task, replace]     True     True
2  [inspection, complete, replaced, horizontal, s...    False    False

【讨论】：

【解决方案2】：

您可以将一些集合算术与列表推导一起使用，如下所示（请注意，我简化了您的示例以具有更明显的测试用例）：

import pandas as pd

list1 = ['installing', 'replace']
set1 = set(list1)

df = pd.DataFrame({'col1': [['daily', 'ask'], 
                            ['daily', 'replace'],
                            ['installing', 'replace', 'blade']]})

# new1 should be True when the intersection of list1 with the row from col1 is not empty
df['new1'] = [set1.intersection(set(row)) != set() for row in df.col1]

# new2 should be True when list1 is a subset of the row from col1 
df['new2'] = [set1.issubset(set(row)) for row in df.col1]

df

    col1                          new1   new2
0   [daily, ask]                  False  False
1   [daily, replace]              True   False
2   [installing, replace, blade]  True   True

【讨论】：