嗨,我正在尝试将我的 df 转换为二进制并将其存储在一个变量中。
我的_df:
df = pd.DataFrame({'A':[1,2,3],'B':[4,5,6]})
Run Code Online (Sandbox Code Playgroud)
我的代码:
import io
towrite = io.BytesIO()
df.to_excel(towrite) # write to BytesIO buffer
towrite.seek(0) # reset pointer
Run Code Online (Sandbox Code Playgroud)
我正进入(状态 AttributeError: '_io.BytesIO' object has no attribute 'write_cells'
完整追溯:
AttributeError Traceback (most recent call last)
<ipython-input-25-be6ee9d9ede6> in <module>()
1 towrite = io.BytesIO()
----> 2 df.to_excel(towrite) # write to BytesIO buffer
3 towrite.seek(0) # reset pointer
4 encoded = base64.b64encode(towrite.read()) #
C:\ProgramData\Anaconda3\lib\site-packages\pandas\core\frame.py in to_excel(self, excel_writer, sheet_name, na_rep, float_format, columns, header, index, index_label, startrow, startcol, engine, merge_cells, encoding, inf_rep, …Run Code Online (Sandbox Code Playgroud) my_list=["one", "one two", "three"]
Run Code Online (Sandbox Code Playgroud)
我正在使用这个列表生成一个文字云
wordcloud = WordCloud(width = 1000, height = 500).generate(" ".join(my_list))
Run Code Online (Sandbox Code Playgroud)
当我将所有项目转换为字符串时,它正在生成单词云
"one","two","three"
But I want to generate word cloud for the values, "one","one two","three"
Run Code Online (Sandbox Code Playgroud)
帮助我为列表中的项目生成文字云
我有一个pd系列,
s = pd.Series([1, 2, 3, np.nan])
Run Code Online (Sandbox Code Playgroud)
当我做,
s.map('this is a string {}'.format)
[out]
0 this is a string 1.0
1 this is a string 2.0
2 this is a string 3.0
3 this is a string nan
Run Code Online (Sandbox Code Playgroud)
如何使用格式化字符串获得相同的结果?
s.map(f'this is a string {?}') ?
Run Code Online (Sandbox Code Playgroud) df=pd.DataFrame({"A":["one","two","three"],"B":["fopur","give","six"]})
Run Code Online (Sandbox Code Playgroud)
当我做,
df.B.str.contains("six").any()
out[2]=True
Run Code Online (Sandbox Code Playgroud)
当我做,
df.B.str.contains("six)").any()
Run Code Online (Sandbox Code Playgroud)
我收到以下错误,
C:\ProgramData\Anaconda3\lib\sre_parse.py in parse(str, flags, pattern)
868 if source.next is not None:
869 assert source.next == ")"
--> 870 raise source.error("unbalanced parenthesis")
871
872 if flags & SRE_FLAG_DEBUG:
error: unbalanced parenthesis at position 3
Run Code Online (Sandbox Code Playgroud)
请帮忙!
my_list = ["on@3", "two#", "thre%e"]
Run Code Online (Sandbox Code Playgroud)
我的预期输出是
out_list = ["one","two","three"]
Run Code Online (Sandbox Code Playgroud)
我不能简单地申请strip()这些项目,请帮忙。
我有一个带有数据表的word文件(.docx),我正在尝试使用该表创建一个pandas数据框,我使用了docx和pandas模块.但我无法创建数据框.
from docx import Document
document = Document('req.docx')
for table in document.tables:
for row in table.rows:
for cell in row.cells:
print (cell.text)
Run Code Online (Sandbox Code Playgroud)
并尝试将表读为df pd.read_table("path of the file")
我可以逐个单元格读取数据,但我想读取整个表格或任何特定的列.提前致谢
我的 Java 程序员将一个 excel 文件转换为二进制文件并将二进制内容发送给我。
他使用sun.misc.BASE64Encoder和sun.misc.BASE64Decoder()进行编码。
我需要使用 python 将该二进制数据转换为数据帧。
数据看起来像,
UEsDBBQABgAIAAAAIQBi7p1oXgEAAJAEAAATAAgCW0NvbnRlbnRfVHl........
我试过bas64解码器但没有帮助。
我的代码:
import base64
with open('encoded_data.txt','rb') as d:
data=d.read()
print(data)
`UEsDBBQABgAIAAAAIQBi7p1oXgEAAJAEAAATAAgCW0NvbnRlbnRfVHl........`
decrypted=base64.b64decode(data)
print(decrypt)
'PK\x03\x04\x14\x00\x06\x00\x08\x00\x00\x00!\x00b\xee\x9dh^\x01\x00\x00\x90\x04\x00\x00\x13\x00\x08\x02[Content_Types].xml \xa2\x04\x02(\xa0\x00\x02\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00
Run Code Online (Sandbox Code Playgroud)
请帮我将此二进制数据转换为熊猫数据框。
我有一个如下所示的 dash.datatable,但列宽对于我的值来说太长,列宽应限制为相应列的最大长度字。
我的代码:
app = JupyterDash(__name__)
df = pd.DataFrame(
{
"A" : ["one","two","three"],
"B" : ["one","two","three"],
"C" : ["one","two","three"],
}
)
app.layout = dash_table.DataTable(
data=df.to_dict('records'),
columns=[{'id': c, 'name': c} for c in df.columns],
style_data={
'whiteSpace': 'normal',
'height': 'auto',
},
)
if __name__ == "__main__":
app.run_server(mode="jupyterlab")
Run Code Online (Sandbox Code Playgroud)
我的数据表看起来像:
我有一个 df,
Name Description
Ram Ram is one of the good cricketer
Sri Sri is one of the member
Kumar Kumar is a keeper
Run Code Online (Sandbox Code Playgroud)
和一个列表,my_list=["one","good","ravi","ball"]
我试图从 my_list 中获取至少有一个关键字的行。
我试过,
mask=df["Description"].str.contains("|".join(my_list),na=False)
Run Code Online (Sandbox Code Playgroud)
我得到了 output_df,
Name Description
Ram Ram is one of ONe crickete
Sri Sri is one of the member
Ravi Ravi is a player, ravi is playing
Kumar there is a BALL
Run Code Online (Sandbox Code Playgroud)
我还想添加“描述”中存在的关键字及其在单独列中的计数,
我想要的输出是,
Name Description pre-keys keys count
Ram Ram is one of ONe crickete one,good,ONe one,good 2
Sri Sri …Run Code Online (Sandbox Code Playgroud) 嗨,我想获取包含特定单词的数据框的列名,
例如:我有一个数据帧,
NA good employee
Not available best employer
not required well manager
not eligible super reportee
my_word=["well"]
Run Code Online (Sandbox Code Playgroud)
如何检查df中是否存在"井",以及"井"中的列名称是否存在
提前致谢!