檢查串列中的字串是否在串列中的其他兩個字串之間？

import pandas as pd
    
nameBank = ["John Doe", "Jane Doe", "Patrick Star", "Spongebob Squarepants"]
phoneList = []
nameList = []

list1 = ["1234567890", "John doe", "Not a NAME/USELESS FILLERINFO",  "2345678901", "jane doe", "Not a NAME/USELESS FILLERINFO", "Not a NAME/USELESS FILLERINFO", "3456789012", "4567890123", "5678901234", "patrick star", "6789012345"]

df = pd.DataFrame({'Phone Number': phoneList, 'Name': nameList})
df.to_csv('results.csv', index=False, encoding='utf-8')
print(df)

我想要做的是從這個 list1 中檢索每個電話號碼并將其放入phoneList.

從那里我想查看nameBank串列中當前電話號碼之后和串列中下一個電話號碼之前的名稱。

如果電話號碼后面有一個名字，那么我希望能夠將它nameList附加到nameList. 所以它基本上可以對應一個excel圖表。

即電話號碼1234567890在兩個串列之間具有與其對應的姓名 John Doe。第二個電話號碼附加了 Jane Doe 的名字，因此當您使用這兩個串列創建使用 Pandas 的表時，它們將對應。第三個電話號碼3456789012與串列中的下一個電話號碼之間沒有名稱，因此我希望 nameList 的附加值是"no name found".

基本上輸出表的樣子：

uj5u.com熱心網友回復：

所以，你想把 list1 決議成一個系列：

list1 = ["1234567890", "John doe", "Not a NAME/USELESS FILLERINFO",  "2345678901", "jane doe", "Not a NAME/USELESS FILLERINFO", "Not a NAME/USELESS FILLERINFO", "3456789012", "4567890123", "5678901234", "patrick star", "6789012345"]


import re

num = re.compile('\d{10}')
output = {}
i = 0
while i < len(list1):
    if not num.match(list1[i]):
        i  = 1
        continue
    output[list1[i]] = list1[i 1] if i 1<len(list1) and not num.match(list1[i 1]) else 'not found'
    i  = 1
    
series = pd.Series(output)

輸出：

1234567890        John doe
2345678901        jane doe
3456789012       not found
4567890123       not found
5678901234    patrick star
6789012345       not found
dtype: object

uj5u.com熱心網友回復：

import pandas as pd

nameBank = ["John Doe", "Jane Doe", "Patrick Star", "Spongebob Squarepants"]
list1 = ["1234567890", "John doe", "Not a NAME/USELESS FILLERINFO",  "2345678901", "jane doe", "Not a NAME/USELESS FILLERINFO", "Not a NAME/USELESS FILLERINFO", "3456789012", "4567890123", "5678901234", "patrick star", "6789012345"]

data = []
for index, elem in enumerate(list1):
    if elem.isnumeric():
        if (len(list1) - 1) > index:
            if list1[index 1].casefold() in map(str.casefold, nameBank):
                data.append([elem,list1[index 1].title()])
            else:
                data.append([elem, 'No Name Found'])
        else:
            data.append([elem, 'No Name Found'])
 
df = pd.DataFrame(data, columns=['Phone Number', 'Name'])      
# df.to_csv('results.csv', index=False, encoding='utf-8'     
print(df)

輸出：

  Phone Number           Name
0   1234567890       John Doe
1   2345678901       Jane Doe
2   3456789012  No Name Found
3   4567890123  No Name Found
4   5678901234   Patrick Star
5   6789012345  No Name Found

uj5u.com熱心網友回復：

import re
import pandas as pd

list1 = ["1234567890", "John doe", "Not a NAME/USELESS FILLERINFO",  "2345678901", "jane doe", "Not a NAME/USELESS FILLERINFO", "Not a NAME/USELESS FILLERINFO", "3456789012", "4567890123", "5678901234", "patrick star", "6789012345"]
nameBank = ["John Doe", "Jane Doe", "Patrick Star", "Spongebob Squarepants"]

def mapList(list1):
    output = []
    for index, item in enumerate(list1, start=0): 
        if re.match("^\d{10}", item):
            # Use any one condition             
            # if index < len(list1) - 1 and list1[index   1] in nameBank:
            if index < len(list1) - 1 and not re.match("^\d{10}", list1[index   1]):
                output.append([list1[index], list1[index 1]]);
            else:
                output.append([list1[index],'No Name Found']);
    return output;


df = pd.DataFrame(mapList(list1), columns=['Phone Number', 'Name'])      
print(df)

輸出：

  Phone Number           Name
0   1234567890       John doe
1   2345678901       jane doe
2   3456789012  No Name Found
3   4567890123  No Name Found
4   5678901234   patrick star
5   6789012345  No Name Found

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