pandas 在熊猫中按位置或索引访问列
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Access column by position or index in pandas
提问by jezrael
I have a list as follows and I search it in a csv file to get the item code associate with it. E.g., for 0 -> item code is 11nm
我有一个列表如下，我在一个 csv 文件中搜索它以获取与其关联的项目代码。例如，对于 0 -> 项目代码是 11nm
L = [0, 2]
CSV file:
0, 11nm
1, 22nm
2, 33nm
3, 44nm
I am currently doing it as follows.
我目前正在这样做。
df = pd.read_csv('item_code.csv', sep = ',')
item_codes= df[df["No"].isin(L)]["item_code"].tolist()
However, now I want to know how to do the same thing for a csv file when the file headings (No, item_code) is unavailable.
但是，现在我想知道当文件标题（否，item_code）不可用时如何对 csv 文件执行相同的操作。
Please help me.
请帮我。
回答by cs95
When the column names are unavailable, you can refer to them by index using df.iloc:
当列名不可用时，您可以使用df.iloc以下方法通过索引引用它们：
item_codes = df[df.iloc[:, 0].isin(L)].iloc[:, 1].tolist()
MCVE:
MCVE：
import pandas as pd
import numpy as np
import io
text = \
'''0, 11nm
1, 22nm
2, 33nm
3, 44nm'''
buf = io.StringIO(text)
df = pd.read_csv(buf, sep=',\s*', header=None, engine='python') # no column names
print(df)
0 1
0 0 11nm
1 1 22nm
2 2 33nm
3 3 44nm
L = [0, 2]
item_codes = df[df.iloc[:, 0].isin(L)].iloc[:, 1]
print(item_codes)
0 11nm
2 33nm
Name: 1, dtype: object
print(item_codes.tolist())
['11nm', '33nm']
Notes:
笔记：
sep=',\s*'is a regex pattern (to specify column delimiters)header=Nonewill prevent any rows from being assignedengine='python'to select the regex engine
sep=',\s*'是正则表达式模式（用于指定列分隔符）header=None将阻止分配任何行engine='python'选择正则表达式引擎
回答by jezrael
You can use parameter namesfor specify columns names, for select column use loc:
您可以使用参数names来指定列名称，用于选择列loc：
df = pd.read_csv('item_code.csv', names=['No','item_code'])
print (df)
No item_code
0 0 11nm
1 1 22nm
2 2 33nm
3 3 44nm
item_codes= df.loc[df["No"].isin(L), "item_code"].tolist()
print (item_codes)
['11nm', '33nm']
Or use parameter header=Nonefor default columns names 0,1...:
或使用参数header=None作为默认列名0,1...：
df = pd.read_csv('item_code.csv', header=None)
print (df)
0 1
0 0 11nm
1 1 22nm
2 2 33nm
3 3 44nm
#first column selected by position with iloc
item_codes= df.loc[df.iloc[:,0].isin(L), 1].tolist()
print (item_codes)
['11nm', '33nm']
#first column selected by column name
item_codes= df.loc[df[0].isin(L), 1].tolist()
print (item_codes)
['11nm', '33nm']
回答by Mohamed Ali JAMAOUI
After reading the csv file with header=None, to let pandas know that you don't have a header in your file:
使用 阅读 csv 文件后header=None，让 Pandas 知道您的文件中没有标题：
df = pd.read_csv('item_code.csv', sep = ',', header=None)
You can use the column index instead of the column name.
您可以使用列索引代替列名。
Like this :
像这样 ：
df[df[0].isin(L)][1].tolist()
or this :
或这个 ：
df[df.iloc[:,0].isin(L)][1].tolist()
Explanation:
解释：
if you print the dataframe after reading it without header with print(df)
如果在没有标题的情况下阅读数据帧后打印数据帧 print(df)
0 1
0 0 11nm
1 1 22nm
2 2 33nm
3 3 44nm
You will notice that pandas assigns the number [0,1]to the column names instead of the ["No", "item_code"]that weren't present as a header. Thus, you can reference each column with its index like this df[0]or df.iloc[:, 0].
您会注意到，pandas 将编号分配[0,1]给列名，而不是["No", "item_code"]未作为标题出现的列名。因此，您可以像这样df[0]或df.iloc[:, 0].
The latter df.iloc[:, 0]tells pandas to take all rows and only column 0.
后者df.iloc[:, 0]告诉 Pandas 获取所有行且仅获取 column 0。
