I’m trying to get the number of rows of dataframe df with Pandas, and here is my code.

Method 1:

total_rows = df.count
print total_rows +1

Method 2:

total_rows = df['First_columnn_label'].count
print total_rows +1

Both the code snippets give me this error:

TypeError: unsupported operand type(s) for +: ‘instancemethod’ and ‘int’

What am I doing wrong?

You can use the .shape property or just len(DataFrame.index). However, there are notable performance differences ( len(DataFrame.index) is fastest):

In [1]: import numpy as np

In [2]: import pandas as pd

In [3]: df = pd.DataFrame(np.arange(12).reshape(4,3))

In [4]: df
Out[4]: 
   0  1  2
0  0  1  2
1  3  4  5
2  6  7  8
3  9  10 11

In [5]: df.shape
Out[5]: (4, 3)

In [6]: timeit df.shape
2.77 µs ± 644 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [7]: timeit df[0].count()
348 µs ± 1.31 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [8]: len(df.index)
Out[8]: 4

In [9]: timeit len(df.index)
990 ns ± 4.97 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

EDIT: As @Dan Allen noted in the comments len(df.index) and df[0].count() are not interchangeable as count excludes NaNs,

Suppose df is your dataframe then:

count_row = df.shape[0]  # gives number of row count
count_col = df.shape[1]  # gives number of col count

Or, more succinctly,

r, c = df.shape

Use len(df). This works as of pandas 0.11 or maybe even earlier.

__len__() is currently (0.12) documented with Returns length of index. Timing info, set up the same way as in root’s answer:

In [7]: timeit len(df.index)
1000000 loops, best of 3: 248 ns per loop

In [8]: timeit len(df)
1000000 loops, best of 3: 573 ns per loop

Due to one additional function call it is a bit slower than calling len(df.index) directly, but this should not play any role in most use cases.

How do I get the row count of a pandas DataFrame?

This table summarises the different situations in which you’d want to count something in a DataFrame (or Series, for completeness), along with the recommended method(s).

Footnotes

1. DataFrame.count returns counts for each column as a Series since the non-null count varies by column.
2. DataFrameGroupBy.size returns a Series, since all columns in the same group share the same row-count.
3. DataFrameGroupBy.count returns a DataFrame, since the non-null count could differ across columns in the same group. To get the group-wise non-null count for a specific column, use df.groupby(...)['x'].count() where “x” is the column to count.

Minimal Code Examples

Below, I show examples of each of the methods described in the table above. First, the setup –

df = pd.DataFrame({
    'A': list('aabbc'), 'B': ['x', 'x', np.nan, 'x', np.nan]})
s = df['B'].copy()

df

   A    B
0  a    x
1  a    x
2  b  NaN
3  b    x
4  c  NaN

s

0      x
1      x
2    NaN
3      x
4    NaN
Name: B, dtype: object

Row Count of a DataFrame: len(df), df.shape[0], or len(df.index)

len(df)
# 5

df.shape[0]
# 5

len(df.index)
# 5

It seems silly to compare the performance of constant time operations, especially when the difference is on the level of “seriously, don’t worry about it”. But this seems to be a trend with other answers, so I’m doing the same for completeness.

Of the 3 methods above, len(df.index) (as mentioned in other answers) is the fastest.

Note

• All the methods above are constant time operations as they are simple attribute lookups.
• df.shape (similar to ndarray.shape) is an attribute that returns a tuple of (# Rows, # Cols). For example, df.shape returns (8, 2) for the example here.

Column Count of a DataFrame: df.shape[1], len(df.columns)

df.shape[1]
# 2

len(df.columns)
# 2

Analogous to len(df.index), len(df.columns) is the faster of the two methods (but takes more characters to type).

Row Count of a Series: len(s), s.size, len(s.index)

len(s)
# 5

s.size
# 5

len(s.index)
# 5

s.size and len(s.index) are about the same in terms of speed. But I recommend len(df).

Note
size is an attribute, and it returns the number of elements (=count of rows for any Series). DataFrames also define a size attribute which returns the same result as df.shape[0] * df.shape[1].

Non-Null Row Count: DataFrame.count and Series.count

The methods described here only count non-null values (meaning NaNs are ignored).

Calling DataFrame.count will return non-NaN counts for each column:

df.count()

A    5
B    3
dtype: int64

For Series, use Series.count to similar effect:

s.count()
# 3

Group-wise Row Count: GroupBy.size

For DataFrames, use DataFrameGroupBy.size to count the number of rows per group.

df.groupby('A').size()

A
a    2
b    2
c    1
dtype: int64

Similarly, for Series, you’ll use SeriesGroupBy.size.

s.groupby(df.A).size()

A
a    2
b    2
c    1
Name: B, dtype: int64

In both cases, a Series is returned. This makes sense for DataFrames as well since all groups share the same row-count.

Group-wise Non-Null Row Count: GroupBy.count

Similar to above, but use GroupBy.count, not GroupBy.size. Note that size always returns a Series, while count returns a Series if called on a specific column, or else a DataFrame.

The following methods return the same thing:

df.groupby('A')['B'].size()
df.groupby('A').size()

A
a    2
b    2
c    1
Name: B, dtype: int64

Meanwhile, for count, we have

df.groupby('A').count()

   B
A   
a  2
b  1
c  0

…called on the entire GroupBy object, v/s,

df.groupby('A')['B'].count()

A
a    2
b    1
c    0
Name: B, dtype: int64

Called on a specific column.

TL;DR

use len(df)

len() is your friend, it can be used for row counts as len(df).

Alternatively, you can access all rows by df.index and all columns by df.columns, and as you can use the len(anyList) for getting the count of list, use len(df.index) for getting the number of rows, and len(df.columns) for the column count.

Or, you can use df.shape which returns the number of rows and columns together, if you want to access the number of rows only use df.shape[0] and for the number of columns only use: df.shape[1].

Apart from above answers use can use df.axes to get the tuple with row and column indexes and then use len() function:

total_rows=len(df.axes[0])
total_cols=len(df.axes[1])

…building on Jan-Philip Gehrcke’s answer.

The reason why len(df) or len(df.index) is faster than df.shape[0]. Look at the code. df.shape is a @property that runs a DataFrame method calling len twice.

df.shape??
Type:        property
String form: <property object at 0x1127b33c0>
Source:     
# df.shape.fget
@property
def shape(self):
    """
    Return a tuple representing the dimensionality of the DataFrame.
    """
    return len(self.index), len(self.columns)

And beneath the hood of len(df)

df.__len__??
Signature: df.__len__()
Source:   
    def __len__(self):
        """Returns length of info axis, but here we use the index """
        return len(self.index)
File:      ~/miniconda2/lib/python2.7/site-packages/pandas/core/frame.py
Type:      instancemethod

len(df.index) will be slightly faster than len(df) since it has one less function call, but this is always faster than df.shape[0]

I come to pandas from R background, and I see that pandas is more complicated when it comes to selecting row or column. I had to wrestle with it for a while, then I found some ways to deal with:

getting the number of columns:

len(df.columns)  
## Here:
#df is your data.frame
#df.columns return a string, it contains column's titles of the df. 
#Then, "len()" gets the length of it.

getting the number of rows:

len(df.index) #It's similar.

In case you want to get the row count in the middle of a chained operation, you can use:

df.pipe(len)

Example:

row_count = (
      pd.DataFrame(np.random.rand(3,4))
      .reset_index()
      .pipe(len)
)

This can be useful if you don’t want to put a long statement inside a len() function.

You could use __len__() instead but __len__() looks a bit weird.

Hey you can use do this also:

Let say df is your dataframe. Then df.shape gives you the shape of your dataframe i.e (row,col)

Thus, assign below command to get the required

 row = df.shape[0], col = df.shape[1]

For dataframe df, a printed comma formatted row count used while exploring data:

def nrow(df):
    print("{:,}".format(df.shape[0]))

Example:

nrow(my_df)
12,456,789

An alternative method to finding out the amount of rows in a dataframe which I think is the most readable variant is pandas.Index.size.

Do note that as I commented on the accepted answer:

Suspected pandas.Index.size would actually be faster than len(df.index) but timeit on my computer tells me otherwise (~150 ns slower per loop).

I’m not sure if this would work(data COULD be omitted), but this may work:

*dataframe name*.tails(1)

and then using this, you could find the number of rows by running the code snippet and looking at the row number that was given to you.

Either of this can do (df is the name of the DataFrame):

Method 1: Using len function:

len(df) will give the number of rows in a DataFrame named df.

Method 2: using count function:

df[col].count() will count the number of rows in a given column col.

df.count() will give the number of rows for all the columns.

I have the following DataFrame (df):

import numpy as np
import pandas as pd

df = pd.DataFrame(np.random.rand(10, 5))

I add more column(s) by assignment:

df['mean'] = df.mean(1)

How can I move the column mean to the front, i.e. set it as first column leaving the order of the other columns untouched?

One easy way would be to reassign the dataframe with a list of the columns, rearranged as needed.

This is what you have now:

In [6]: df
Out[6]:
          0         1         2         3         4      mean
0  0.445598  0.173835  0.343415  0.682252  0.582616  0.445543
1  0.881592  0.696942  0.702232  0.696724  0.373551  0.670208
2  0.662527  0.955193  0.131016  0.609548  0.804694  0.632596
3  0.260919  0.783467  0.593433  0.033426  0.512019  0.436653
4  0.131842  0.799367  0.182828  0.683330  0.019485  0.363371
5  0.498784  0.873495  0.383811  0.699289  0.480447  0.587165
6  0.388771  0.395757  0.745237  0.628406  0.784473  0.588529
7  0.147986  0.459451  0.310961  0.706435  0.100914  0.345149
8  0.394947  0.863494  0.585030  0.565944  0.356561  0.553195
9  0.689260  0.865243  0.136481  0.386582  0.730399  0.561593

In [7]: cols = df.columns.tolist()

In [8]: cols
Out[8]: [0L, 1L, 2L, 3L, 4L, 'mean']

Rearrange cols in any way you want. This is how I moved the last element to the first position:

In [12]: cols = cols[-1:] + cols[:-1]

In [13]: cols
Out[13]: ['mean', 0L, 1L, 2L, 3L, 4L]

Then reorder the dataframe like this:

In [16]: df = df[cols]  #    OR    df = df.ix[:, cols]

In [17]: df
Out[17]:
       mean         0         1         2         3         4
0  0.445543  0.445598  0.173835  0.343415  0.682252  0.582616
1  0.670208  0.881592  0.696942  0.702232  0.696724  0.373551
2  0.632596  0.662527  0.955193  0.131016  0.609548  0.804694
3  0.436653  0.260919  0.783467  0.593433  0.033426  0.512019
4  0.363371  0.131842  0.799367  0.182828  0.683330  0.019485
5  0.587165  0.498784  0.873495  0.383811  0.699289  0.480447
6  0.588529  0.388771  0.395757  0.745237  0.628406  0.784473
7  0.345149  0.147986  0.459451  0.310961  0.706435  0.100914
8  0.553195  0.394947  0.863494  0.585030  0.565944  0.356561
9  0.561593  0.689260  0.865243  0.136481  0.386582  0.730399

You could also do something like this:

df = df[['mean', '0', '1', '2', '3']]

You can get the list of columns with:

cols = list(df.columns.values)

The output will produce:

['0', '1', '2', '3', 'mean']

…which is then easy to rearrange manually before dropping it into the first function

Just assign the column names in the order you want them:

In [39]: df
Out[39]: 
          0         1         2         3         4  mean
0  0.172742  0.915661  0.043387  0.712833  0.190717     1
1  0.128186  0.424771  0.590779  0.771080  0.617472     1
2  0.125709  0.085894  0.989798  0.829491  0.155563     1
3  0.742578  0.104061  0.299708  0.616751  0.951802     1
4  0.721118  0.528156  0.421360  0.105886  0.322311     1
5  0.900878  0.082047  0.224656  0.195162  0.736652     1
6  0.897832  0.558108  0.318016  0.586563  0.507564     1
7  0.027178  0.375183  0.930248  0.921786  0.337060     1
8  0.763028  0.182905  0.931756  0.110675  0.423398     1
9  0.848996  0.310562  0.140873  0.304561  0.417808     1

In [40]: df = df[['mean', 4,3,2,1]]

Now, ‘mean’ column comes out in the front:

In [41]: df
Out[41]: 
   mean         4         3         2         1
0     1  0.190717  0.712833  0.043387  0.915661
1     1  0.617472  0.771080  0.590779  0.424771
2     1  0.155563  0.829491  0.989798  0.085894
3     1  0.951802  0.616751  0.299708  0.104061
4     1  0.322311  0.105886  0.421360  0.528156
5     1  0.736652  0.195162  0.224656  0.082047
6     1  0.507564  0.586563  0.318016  0.558108
7     1  0.337060  0.921786  0.930248  0.375183
8     1  0.423398  0.110675  0.931756  0.182905
9     1  0.417808  0.304561  0.140873  0.310562

How about:

df.insert(0, 'mean', df.mean(1))

http://pandas.pydata.org/pandas-docs/stable/dsintro.html#column-selection-addition-deletion

In your case,

df = df.reindex(columns=['mean',0,1,2,3,4])

will do exactly what you want.

In my case (general form):

df = df.reindex(columns=sorted(df.columns))
df = df.reindex(columns=(['opened'] + list([a for a in df.columns if a != 'opened']) ))

You need to create a new list of your columns in the desired order, then use df = df[cols] to rearrange the columns in this new order.

cols = ['mean']  + [col for col in df if col != 'mean']
df = df[cols]

You can also use a more general approach. In this example, the last column (indicated by -1) is inserted as the first column.

cols = [df.columns[-1]] + [col for col in df if col != df.columns[-1]]
df = df[cols]

You can also use this approach for reordering columns in a desired order if they are present in the DataFrame.

inserted_cols = ['a', 'b', 'c']
cols = ([col for col in inserted_cols if col in df] 
        + [col for col in df if col not in inserted_cols])
df = df[cols]

import numpy as np
import pandas as pd
df = pd.DataFrame()
column_names = ['x','y','z','mean']
for col in column_names: 
    df[col] = np.random.randint(0,100, size=10000)

You can try out the following solutions :

Solution 1:

df = df[ ['mean'] + [ col for col in df.columns if col != 'mean' ] ]

Solution 2:

df = df[['mean', 'x', 'y', 'z']]

Solution 3:

col = df.pop("mean")
df = df.insert(0, col.name, col)

Solution 4:

df.set_index(df.columns[-1], inplace=True)
df.reset_index(inplace=True)

Solution 5:

cols = list(df)
cols = [cols[-1]] + cols[:-1]
df = df[cols]

solution 6:

order = [1,2,3,0] # setting column's order
df = df[[df.columns[i] for i in order]]

Time Comparison:

Solution 1:

CPU times: user 1.05 ms, sys: 35 µs, total: 1.08 ms Wall time: 995 µs

Solution 2:

CPU times: user 933 µs, sys: 0 ns, total: 933 µs Wall time: 800 µs

Solution 3:

CPU times: user 0 ns, sys: 1.35 ms, total: 1.35 ms Wall time: 1.08 ms

Solution 4:

CPU times: user 1.23 ms, sys: 45 µs, total: 1.27 ms Wall time: 986 µs

Solution 5:

CPU times: user 1.09 ms, sys: 19 µs, total: 1.11 ms Wall time: 949 µs

Solution 6:

CPU times: user 955 µs, sys: 34 µs, total: 989 µs Wall time: 859 µs

From August 2018:

If your column names are too long to type then you could specify the new order through a list of integers with the positions:

Data:

          0         1         2         3         4      mean
0  0.397312  0.361846  0.719802  0.575223  0.449205  0.500678
1  0.287256  0.522337  0.992154  0.584221  0.042739  0.485741
2  0.884812  0.464172  0.149296  0.167698  0.793634  0.491923
3  0.656891  0.500179  0.046006  0.862769  0.651065  0.543382
4  0.673702  0.223489  0.438760  0.468954  0.308509  0.422683
5  0.764020  0.093050  0.100932  0.572475  0.416471  0.389390
6  0.259181  0.248186  0.626101  0.556980  0.559413  0.449972
7  0.400591  0.075461  0.096072  0.308755  0.157078  0.207592
8  0.639745  0.368987  0.340573  0.997547  0.011892  0.471749
9  0.050582  0.714160  0.168839  0.899230  0.359690  0.438500

Generic example:

new_order = [3,2,1,4,5,0]
print(df[df.columns[new_order]])  

          3         2         1         4      mean         0
0  0.575223  0.719802  0.361846  0.449205  0.500678  0.397312
1  0.584221  0.992154  0.522337  0.042739  0.485741  0.287256
2  0.167698  0.149296  0.464172  0.793634  0.491923  0.884812
3  0.862769  0.046006  0.500179  0.651065  0.543382  0.656891
4  0.468954  0.438760  0.223489  0.308509  0.422683  0.673702
5  0.572475  0.100932  0.093050  0.416471  0.389390  0.764020
6  0.556980  0.626101  0.248186  0.559413  0.449972  0.259181
7  0.308755  0.096072  0.075461  0.157078  0.207592  0.400591
8  0.997547  0.340573  0.368987  0.011892  0.471749  0.639745
9  0.899230  0.168839  0.714160  0.359690  0.438500  0.050582

And for the specific case of OP’s question:

new_order = [-1,0,1,2,3,4]
df = df[df.columns[new_order]]
print(df)

       mean         0         1         2         3         4
0  0.500678  0.397312  0.361846  0.719802  0.575223  0.449205
1  0.485741  0.287256  0.522337  0.992154  0.584221  0.042739
2  0.491923  0.884812  0.464172  0.149296  0.167698  0.793634
3  0.543382  0.656891  0.500179  0.046006  0.862769  0.651065
4  0.422683  0.673702  0.223489  0.438760  0.468954  0.308509
5  0.389390  0.764020  0.093050  0.100932  0.572475  0.416471
6  0.449972  0.259181  0.248186  0.626101  0.556980  0.559413
7  0.207592  0.400591  0.075461  0.096072  0.308755  0.157078
8  0.471749  0.639745  0.368987  0.340573  0.997547  0.011892
9  0.438500  0.050582  0.714160  0.168839  0.899230  0.359690

The main problem with this approach is that calling the same code multiple times will create different results each time, so one needs to be careful :)

This function avoids you having to list out every variable in your dataset just to order a few of them.

def order(frame,var):
    if type(var) is str:
        var = [var] #let the command take a string or list
    varlist =[w for w in frame.columns if w not in var]
    frame = frame[var+varlist]
    return frame 

It takes two arguments, the first is the dataset, the second are the columns in the data set that you want to bring to the front.

So in my case I have a data set called Frame with variables A1, A2, B1, B2, Total and Date. If I want to bring Total to the front then all I have to do is:

frame = order(frame,['Total'])

If I want to bring Total and Date to the front then I do:

frame = order(frame,['Total','Date'])

EDIT:

Another useful way to use this is, if you have an unfamiliar table and you’re looking with variables with a particular term in them, like VAR1, VAR2,… you may execute something like:

frame = order(frame,[v for v in frame.columns if "VAR" in v])

I ran into a similar question myself, and just wanted to add what I settled on. I liked the reindex_axis() method for changing column order. This worked:

df = df.reindex_axis(['mean'] + list(df.columns[:-1]), axis=1)

An alternate method based on the comment from @Jorge:

df = df.reindex(columns=['mean'] + list(df.columns[:-1]))

Although reindex_axis seems to be slightly faster in micro benchmarks than reindex, I think I prefer the latter for its directness.

Simply do,

df = df[['mean'] + df.columns[:-1].tolist()]

You could do the following (borrowing parts from Aman’s answer):

cols = df.columns.tolist()
cols.insert(0, cols.pop(-1))

cols
>>>['mean', 0L, 1L, 2L, 3L, 4L]

df = df[cols]

Just type the column name you want to change, and set the index for the new location.

def change_column_order(df, col_name, index):
    cols = df.columns.tolist()
    cols.remove(col_name)
    cols.insert(index, col_name)
    return df[cols]

For your case, this would be like:

df = change_column_order(df, 'mean', 0)

Moving any column to any position:

import pandas as pd
df = pd.DataFrame({"A": [1,2,3], 
                   "B": [2,4,8], 
                   "C": [5,5,5]})

cols = df.columns.tolist()
column_to_move = "C"
new_position = 1

cols.insert(new_position, cols.pop(cols.index(column_to_move)))
df = df[cols]

I think this is a slightly neater solution:

df.insert(0,'mean', df.pop("mean"))

This solution is somewhat similar to @JoeHeffer ‘s solution but this is one liner.

Here we remove the column "mean" from the dataframe and attach it to index 0 with the same column name.

Here’s a way to move one existing column that will modify the existing data frame in place.

my_column = df.pop('column name')
df.insert(3, my_column.name, my_column)

This question has been answered before but reindex_axis is deprecated now so I would suggest to use:

df.reindex(sorted(df.columns), axis=1)

How about using “T”?

df.T.reindex(['mean',0,1,2,3,4]).T

@clocker: Your solution was very helpful for me, as I wanted to bring two columns in front from a dataframe where I do not know exactly the names of all columns, because they are generated from a pivot statement before. So, if you are in the same situation: To bring columns in front that you know the name of and then let them follow by “all the other columns”, I came up with the following general solution;

df = df.reindex_axis(['Col1','Col2'] + list(df.columns.drop(['Col1','Col2'])), axis=1)

set():

A simple approach is using set(), in particular when you have a long list of columns and do not want to handle them manually:

cols = list(set(df.columns.tolist()) - set(['mean']))
cols.insert(0, 'mean')
df = df[cols]

I liked Shoresh’s answer to use set functionality to remove columns when you don’t know the location, however this didn’t work for my purpose as I need to keep the original column order (which has arbitrary column labels).

I got this to work though by using IndexedSet from the boltons package.

I also needed to re-add multiple column labels, so for a more general case I used the following code:

from boltons.setutils import IndexedSet
cols = list(IndexedSet(df.columns.tolist()) - set(['mean', 'std']))
cols[0:0] =['mean', 'std']
df = df[cols]

Hope this is useful to anyone searching this thread for a general solution.

You can use reindex which can be used for both axis:

df
#           0         1         2         3         4      mean
# 0  0.943825  0.202490  0.071908  0.452985  0.678397  0.469921
# 1  0.745569  0.103029  0.268984  0.663710  0.037813  0.363821
# 2  0.693016  0.621525  0.031589  0.956703  0.118434  0.484254
# 3  0.284922  0.527293  0.791596  0.243768  0.629102  0.495336
# 4  0.354870  0.113014  0.326395  0.656415  0.172445  0.324628
# 5  0.815584  0.532382  0.195437  0.829670  0.019001  0.478415
# 6  0.944587  0.068690  0.811771  0.006846  0.698785  0.506136
# 7  0.595077  0.437571  0.023520  0.772187  0.862554  0.538182
# 8  0.700771  0.413958  0.097996  0.355228  0.656919  0.444974
# 9  0.263138  0.906283  0.121386  0.624336  0.859904  0.555009

df.reindex(['mean', *range(5)], axis=1)

#        mean         0         1         2         3         4
# 0  0.469921  0.943825  0.202490  0.071908  0.452985  0.678397
# 1  0.363821  0.745569  0.103029  0.268984  0.663710  0.037813
# 2  0.484254  0.693016  0.621525  0.031589  0.956703  0.118434
# 3  0.495336  0.284922  0.527293  0.791596  0.243768  0.629102
# 4  0.324628  0.354870  0.113014  0.326395  0.656415  0.172445
# 5  0.478415  0.815584  0.532382  0.195437  0.829670  0.019001
# 6  0.506136  0.944587  0.068690  0.811771  0.006846  0.698785
# 7  0.538182  0.595077  0.437571  0.023520  0.772187  0.862554
# 8  0.444974  0.700771  0.413958  0.097996  0.355228  0.656919
# 9  0.555009  0.263138  0.906283  0.121386  0.624336  0.859904

Here is a function to do this for any number of columns.

def mean_first(df):
    ncols = df.shape[1]        # Get the number of columns
    index = list(range(ncols)) # Create an index to reorder the columns
    index.insert(0,ncols)      # This puts the last column at the front
    return(df.assign(mean=df.mean(1)).iloc[:,index]) # new df with last column (mean) first