Get row and column in Pandas for a cell with a certain value

Create a df with NaN where your_value is not found.
Drop all rows that don't contain the value.
Drop all columns that don't contain the value

    a = df.where(df=='your_value').dropna(how='all').dropna(axis=1)

To get the row(s)

    a.index

To get the column(s)

    a.columns  

You can do some long and hard to read list comprehension:

# assume this df and that we are looking for 'abc'
df = pd.DataFrame({'col':['abc', 'def','wert','abc'], 'col2':['asdf', 'abc', 'sdfg', 'def']})

[(df[col][df[col].eq('abc')].index[i], df.columns.get_loc(col)) for col in df.columns for i in range(len(df[col][df[col].eq('abc')].index))]

out:

[(0, 0), (3, 0), (1, 1)]

I should note that this is (index value, column location)

you can also change .eq() to str.contains() if you are looking for any strings that contains a certain value:

[(df[col][df[col].str.contains('ab')].index[i], df.columns.get_loc(col)) for col in df.columns for i in range(len(df[col][df[col].str.contains('ab')].index))]

You can simply create a mask of the same shape than your df by calling df == 'title'. You can then combines this with the df.where() method, which will set all fields to NA that are different to your keyword, and finally you can use dropna() to reduce it to all valid fields. Then you can use the df.columnnsand df.indexlike you're use to.

df = pd.DataFrame({"a": [0,1,2], "b": [0, 9, 7]})
print(df.where(df == 0).dropna().index)
print(df.where(df == 0).dropna().columns)

#Int64Index([0], dtype='int64')
#Index(['a', 'b'], dtype='object')

Here's an example to fetch all the row and column index of the cells containing word 'title' -

df = pd.DataFrame({'A':['here goes the title', 'tt', 'we have title here'],
                  'B': ['ty', 'title', 'complex']})
df


+---+---------------------+---------+
|   |          A          |    B    |
+---+---------------------+---------+
| 0 | here goes the title | ty      |
| 1 | tt                  | title   |
| 2 | we have title here  | complex |
+---+---------------------+---------+


idx = df.apply(lambda x: x.str.contains('title'))

col_idx = []
for i in range(df.shape[1]):
    col_idx.append(df.iloc[:,i][idx.iloc[:,i]].index.tolist())


out = []
cnt = 0
for i in col_idx:
    for j in range(len(i)):
        out.append((i[j], cnt))
    cnt += 1
out

# [(0, 0), (2, 0), (1, 1)]   # Expected output

Another approach that's in the vein of @It_is_Chris's solution, but may be a little easier to read:

# assuming this df and that we are looking for 'abc'
df = pd.DataFrame({'col':['abc', 'def','wert','abc'], 'col2':['asdf', 'abc', 'sdfg', 'def']})
[x[1:] for x in ((v, i, j) for i, row_tup in enumerate(df.itertuples(index=False)) for j, v in enumerate(row_tup)) if x[0] == "abc"]

Output

[(0, 0), (1, 1), (3, 0)]

The answer by @firefly works if the second dropna gets the how='all', too like so:

a = df.where(targetMap == 'your_value').dropna(how='all').dropna(how='all',axis=1)

Similar to what Chris said, I found this to work for me, although it's not the prettiest or shortest way. This returns all the row,column pairs matching a regular expression in a dataframe:

for row in df.itertuples():
    col_count = 0
    for col in row:
        if regex.match(str(col)):
            tuples.append((row_count, col_count))
            col_count+=1
        row_count+=1

return tuples