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从给定熊猫数据框的列名中获取列索引

原文:https://www.geesforgeks.org/get-column-index-from-column-给定熊猫的名字-dataframe/

在本文中,我们将看到如何从数据帧的列名中获取列索引。我们将使用data frame.columns属性和index.get_loc方法的 熊猫 模块结合在一起。****

**语法:数据帧.列****

**返回:列名索引****

**语法: Index.get_loc(键,方法=无,容差=无)

返回: loc:如果索引唯一,则为 int,如果索引单调,则为 slice,否则为 mask****

**代码:让我们创建一个数据帧:****

*蟒蛇 3*

**# import pandas library
import pandas as pd

# dictionary
record = {'Math': [10, 20, 30, 
                   40, 70],
          'Science': [40, 50, 60,
                      90, 50], 
          'English': [70, 80, 66, 
                      75, 88]}

# create a dataframe
df = pd.DataFrame(record)

# show the dataframe
print(df)**

**输出:****

****Dataframe

数据帧****

**例 1: 获取“科学”一栏的索引号。****

*蟒蛇 3*

**# import pandas library
import pandas as pd

# dictionary
record = {'Math': [10, 20, 30, 40, 70],
          'Science': [40, 50, 60, 90, 50], 
          'English': [70, 80, 66, 75, 88]}

# give column name
col_name = "Science"

# find the index no
index_no = df.columns.get_loc(col_name)

print("Index of {} column in given dataframe is : {}".format(col_name, index_no))**

**输出:****

****index number of Science column

科学专栏索引号****

**例 2: 获取“英语”列的索引号。****

*蟒蛇 3*

**# import pandas library
import pandas as pd

# dictionary
record = {'Math': [10, 20, 30,
                   40, 70],
          'Science': [40, 50, 60,
                      90, 50], 
          'English': [70, 80, 66,
                      75, 88]}

# create a dataframe
df = pd.DataFrame(record)

# give column name
col_name = "English"

# find the index no
index_no = df.columns.get_loc(col_name)

print("Index of {} column in given dataframe is : {}".format(col_name, index_no))**

**输出:****

****index number of English column

英文栏目索引号****



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