将Python中的一列拆分为多个列

Question

我有一些非常简单的代码输出：

Name
Workplace
And a abstract

然后，这一过程一遍又一遍地重复。所以：

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

我需要把它分成三列：

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，它会循环回到开始。但是，我不显示此标记。因此，一个记录是名称、工作场所和摘要，直到它遇到一个新的<h1>标记。你知道吗

以下是我的代码：

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

我正在努力把它分开，并把列。你知道吗

Answer 1

如果您想处理自己的输入格式，您需要 一些假设。对于这个代码示例，我假设“h1”出现在三行集合之间。如果中间允许，代码需要稍微不同。你知道吗

想法：

• 编写一个生成器函数，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，将结果移到pandas数据框中

这是一个有效的例子。你知道吗

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)

数据帧打印为

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

如果您需要像您的示例一样打印数据帧， 下面是示例代码。你知道吗

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python3.6，如果您使用的是旧版本，则需要更改print命令。你知道吗

相比之下，使用Pandas可以这样做（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]))

输出相同的结果。你知道吗

NameCol WorkplaceCol    AbstractCol

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

Answer 2

假设你有这样一个df：

col1
NameA
WorkplaceA
AbstractA
NameB
WorkplaceB
AbstractB

你可以做：

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)

它会给你：

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