Question

我有一些代码可以很简单地输出：

Name
Workplace
And a abstract

然后重复一遍又一遍。所以：

NameA
WorkplaceA
And a abstractA
NameB
WorkplaceB
And a abstractB
etc...

我需要将其分为三列：

NameCol  WorkplaceCol  AbstractCol

NameA    WorkplaceA    AbstractA
NameB    WorkplaceB    AbstractB
NameC    WorkplaceC    AbstractC
etc...

当我的代码找到一个<h1>标记时，它循环回到开始。但是，我不显示此标签。因此，一条记录是名称，工作场所和摘要，直到遇到新的<h1>标签为止。

这是我的代码：

headernum = 0
i = 0
x = soup.find_all("h1")

for i in range(len(x)):
    header = soup.find_all('h1')[headernum]
    name = header.find_all_next('p')[1]
    print(name.text)
    workplace = name.find_all_next('i')[0]
    print(workplace.text)
    abstract = []
    for elem in name.next_siblings:
        if elem.name == 'h1':
            break
        if elem.name != 'p':
            continue
        abstract.append(elem.get_text())
    x = " ".join(abstract).replace("\n", " ").encode('utf-8')
    print(x)
    i += 1
    headernum += 1

我正在努力将其拆分并放入列中。

Answer 1

假设您拥有这样的df：

col1
NameA
WorkplaceA
AbstractA
NameB
WorkplaceB
AbstractB

您可以：

import numpy as np

# Set the same number for each 3 lines
df['index'] = df.index / 3
df['index'] = df['index'].apply(np.floor)

# Set 0 for Names, 1 for Workplaces and 2 for Abstract
df["type_id"] = df.index % 3

# Rename 0, 1 and 2 by a label
df["type_label"] = df["type_id"].map({0: "Name", 1: "Workplace", 2: "Abstract"})

# Pivot the table
df = df.pivot(index='index', columns='type_label', values='col1')
print(df)

它将给您：

type_label   Abstract   Name   Workplace
index
0.0         AbstractA  NameA  WorkplaceA
1.0         AbstractB  NameB  WorkplaceB

Answer 2

如果要处理自己的输入格式，则需要一些假设。对于此代码示例，我假设“ h1”出现在三行之间。如果中间允许，则代码需要稍有不同。

想法：

编写一个生成器函数，该函数循环遍历文本并以字典形式返回每一整行。
全部收集
当您将问题标记为“ pandas”时，将结果移至pandas数据框

这是一个可行的示例。

import pandas as pd

example_text="""NameA
WorkplaceA
And a abstractA
NameB
WorkplaceB
And a abstractB
<h1>
NameC
WorkplaceC
And a abstractC"""

def next_name(mystr):
    lines = iter(mystr.split('\n'))
    while True:
        n = {'NameCol':None,
         'WorkplaceCol':None,
         'AbstractCol':None
        }
        try:
            n['NameCol'] = next(lines)
            if n['NameCol'] == '<h1>':
                continue
            n['WorkplaceCol'] = next(lines)
            if n['WorkplaceCol'] == '<h1>':
                continue
            n['AbstractCol'] = next(lines)
            if n['AbstractCol'] == '<h1>':
                continue
            yield n 
        except StopIteration:
            break

df = pd.DataFrame(next_name(example_text), columns=['NameCol','WorkplaceCol','AbstractCol'])
print(df)

数据框打印为

  NameCol WorkplaceCol      AbstractCol
0   NameA   WorkplaceA  And a abstractA
1   NameB   WorkplaceB  And a abstractB
2   NameC   WorkplaceC  And a abstractC

如果您需要完全按照示例打印数据框，这是示例代码。

print(''.join(f'{x}\t' for x in df.columns))
print()
for row in df.iterrows():
    print(''.join(f'{x}\t' for x in row[1]))

输出

NameCol WorkplaceCol    AbstractCol 

NameA   WorkplaceA  And a abstractA 
NameB   WorkplaceB  And a abstractB 
NameC   WorkplaceC  And a abstractC

注意：我使用的是Python 3.6，如果您使用的是旧版本，则需要更改print命令。

相比之下，使用Pandas可以做到这一点（使用上面代码中的示例）

df = pd.DataFrame(example_text.split('\n'))
df = df[df[0] != '<h1>'].reset_index().copy()
df['row'] = df.index // 3
result = df.groupby('row').agg(lambda x: list(x))[0].values

print('\t'.join(["NameCol", "WorkplaceCol", "AbstractCol"]))
print('')
print('\n'.join(['\t'.join(x) for x in result]))

输出相同。

NameCol WorkplaceCol    AbstractCol

NameA   WorkplaceA  And a abstractA
NameB   WorkplaceB  And a abstractB
NameC   WorkplaceC  And a abstractC

在Python中将一列拆分为多列

2 个答案: