如何在 Pandas 中获取 dataframe 中的行号和列号？

Question

How can I get the number of the row and the column in a dataframe that contains a certain value using Pandas? For example, I have the following dataframe:

For example, i need to know the row and column of "Smith" (row 1, column LastName)

Answer 1

Maybe this is a solution or a first step to a solution.

If you filter for the value you are looking for all items which are not the value you want are replaced with NaN . Now you can drop all columns where all values are NaN . This leaves a DataFrame with your item and the indices. Then you can ask for index and name.

import numpy as np
import pandas as pd
df = pd.DataFrame({'LastName':['a', 'Smith', 'b'], 'other':[1,2,3]})

value = df[df=='Smith'].dropna(axis=0, how='all').dropna(axis=1, how='all')
print(value.index.values)
print(value.columns.values)

But I think this can be improved.

Answer 2

Here's a one liner that efficiently gets the row and column of a value:

df = pd.DataFrame({"ClientID": [34, 67, 53], "LastName": ["Johnson", "Smith", "Brows"] })
result = next(x[1:] for x in ((v, i, j) for i, row_tup in enumerate(df.itertuples(index=False)) for j, v in zip(df.columns, row_tup)) if x[0] == "Smith")
print(result)

Output

(1, "LastName")

Unpacking that one liner

# This is a generator that unpacks the dataframe and gets the value, row number (i) and column name (j) for every value in the dataframe
item_generator = ((v, i, j) for i, row_tup in enumerate(df.itertuples(index=False)) for j, v in zip(df.columns, row_tup))
# This iterates through the generator until it finds a match
# It outputs just the row and column number by leaving off the first item in the tuple
next(x[1:] for x in item_generator if x[0] == "Smith")

Props to this this answer for the second half of the solution

Answer 3

Just to add another possible solution to the bucket. If you really need to your search your whole DataFrame, you may consider using numpy.where , such as:

import numpy as np

value = 'Smith'
rows, cols = np.where(df.values == value)

where_are_you = [(df.index[row], df.columns[col]) for row, col in zip(rows, cols)]

So, if your DataFrame is like

   ClientID First Name LastName
0        34         Mr    Smith
1        67      Keanu   Reeves
2        53     Master     Yoda
3        99      Smith    Smith
4       100      Harry   Potter

The code output will be:

[(0, 'LastName'), (3, 'First Name'), (3, 'LastName')]

Edit: Just to satisfy everybody's curiosity, here it is a benchmark of all answers

The code is written below. I removed the print statements to be fair, because they would make codes really slow for bigger dataframes.

val = 0

def setup(n=10):
    return pd.DataFrame(np.random.randint(-100, 100, (n, 3)))


def nested_for(df):
    index = df.index  # Allows to get the row index
    columns = df.columns  # Allows to get the column name
    value_to_be_checked = val
    for i in index[df.isin([value_to_be_checked]).any(axis=1)].to_list():
        for j, e in enumerate(df.iloc[i]):
            if e == value_to_be_checked:
                _ = "(row {}, column {})".format(i, columns[j])


def df_twin_dropna(df):
    value = df[df == val].dropna(axis=0, how='all').dropna(axis=1, how='all')
    return value.index.values, value.columns.values


def numpy_where(df):
    rows, cols = np.where(df.values == val)
    return [(df.index[row], df.columns[col]) for row, col in zip(rows, cols)]


def one_line_generator(df):
    return [x[1:] for x in ((v, i, j) for i, row_tup in enumerate(df.itertuples(index=False))
                            for j, v in zip(df.columns, row_tup)) if x[0] == "Smith"]

Answer 4

I tried to simplify the code and make it more readable. This is my attempt:

df = pd.DataFrame({'points': [25, 12, 15, 14, 19],
                   'assists': [5, 7, 7, 9, 12],
                   'rebounds': [11, 8, 10, 6, 6]})
index = df.index # Allows to get the row index
columns = df.columns # Allows to get the column name
value_to_be_checked = 6
for i in index[df.isin([value_to_be_checked]).any(axis=1)].to_list():
  for j, e in enumerate(df.iloc[i]):
    if e == value_to_be_checked:
      print("(row {}, column {})".format(i, column[j])

Answer 5

You can do this by looping though all the columns and finding the matching rows. This will give you a list of all the cells that matches your criteria:

Method 1(without comprehension):

import pandas as pd
# assume this df and that we are looking for 'abc'
df = pd.DataFrame({
    'clientid': [34, 67, 53],
    'lastname': ['Johnson', 'Smith', 'Brows']
})

Searchval = 'Smith'
l1 = []
#loop though all the columns
for col in df.columns:
    #finding the matching rows
    for i in range(len(df[col][df[col].eq(Searchval)].index)):
        #appending the output to the list
        l1.append((df[col][df[col].eq(Searchval)].index[i], col))
print(l1)

Method 2 (With comprehension):

import pandas as pd

df = pd.DataFrame({
    'clientid': [34, 67, 53],
    'lastname': ['Johnson', 'Smith', 'Brows']
})
#Value to search
Searchval = 'Smith'
#using list comprehension to find the rows in each column which matches the criteria
#and saving it in a list in case we get multiple matches
l = [(df[col][df[col].eq(Searchval)].index[i], col) for col in df.columns
     for i in range(len(df[col][df[col].eq(Searchval)].index))]

print(l)

Answer 6

Thanks for submitting your request. This is something you can find with a Google search. Please make some attempt to find answers before asking a new question.

You can find simple and excellent dataframe examples that include column and row selection here: https://studymachinelearning.com/python-pandas-dataframe/

You can also see the official documentation here: https://pandas.pydata.org/pandas-docs/stable/

Select a column by column name:

df['col']

select a row by index:

df.loc['b']