You are first adding R0Cx and then R1Cxy. You need to add RxCy. So try:

newlist = []
row = A
col = B
for x in range (0, row): 
  newlist.append([])

  for y in range(0, col):
    newlist[x].append('R' + str(x) + 'C' + str(y))

print(newlist)

You need to add the columns one by one.

for col in new_cols:
    df[col] = 0

Also see the answers in here for other methods.

Pandas >= 0.25

Pandas can do this in a single function call via df.explode.

df.explode('column_x')

  column_a column_b column_x
0      a_1      b_1      c_1
0      a_1      b_1      c_2
1      a_2      b_2      d_1
1      a_2      b_2      d_2

Note that you can only explode a Series/DataFrame on one column.

Pandas < 0.25

Call np.repeat along the 0^th axis for every column besides column_x.

df1 = pd.DataFrame(
    df.drop('column_x', 1).values.repeat(df['column_x'].str.len(), axis=0),
    columns=df.columns.difference(['column_x'])
)
df1['column_x'] = np.concatenate(df['column_x'].values)

df1

  column_a column_b column_x
0      a_1      b_1      c_1
1      a_1      b_1      c_2
2      a_2      b_2      d_1
3      a_2      b_2      d_2

df.iloc[:, -6: -1].values.T.ravel()

yields

array([11, 18,  5, 14, 31, 26, 12, 37, 12, 13,  3,  2,  4, 14, 41, 18,  5,
       14, 31, 26, 12, 37, 12, 13,  3,  2,  4, 14, 41, 31, 26, 15, 35, 32,
       20, 49, 20, 15,  7,  5, 10, 18, 58, 39, 27, 32, 56, 44, 25, 50, 27,
       35, 44, 10, 31, 58, 68, 45, 44, 68, 59, 45, 50, 56, 38, 62, 63, 20,
       56, 59, 72, 55, 57, 72, 63, 58, 51, 57, 75, 74, 67, 38, 66, 68, 73,
        6,  7,  8,  6, 11,  7,  8,  4, 12, 12, 14,  7,  4,  1])

You can call the tolist() method on the resulting array if you prefer lists.

Starting from Pandas 0.25.0, there is internal method DataFrame.explode(), which was designed just for that:

res = df.explode("b")

output

In [98]: res
Out[98]:
   a  b
0  1  1
0  1  2
1  2  2
1  2  3
1  2  4
2  3  5

Solution for Pandas versions < 0.25: generic vectorized approach - will work also for multiple columns DFs:

assuming we have the following DF:

In [159]: df
Out[159]:
   a          b  c
0  1     [1, 2]  5
1  2  [2, 3, 4]  6
2  3        [5]  7

Solution:

In [160]: lst_col = 'b'

In [161]: pd.DataFrame({
     ...:     col:np.repeat(df[col].values, df[lst_col].str.len())
     ...:     for col in df.columns.difference([lst_col])
     ...: }).assign(**{lst_col:np.concatenate(df[lst_col].values)})[df.columns.tolist()]
     ...:
Out[161]:
   a  b  c
0  1  1  5
1  1  2  5
2  2  2  6
3  2  3  6
4  2  4  6
5  3  5  7

Setup:

df = pd.DataFrame({
    "a" : [1,2,3],
    "b" : [[1,2],[2,3,4],[5]],
    "c" : [5,6,7]
})

Vectorized NumPy approach:

In [124]: pd.DataFrame({'a':np.repeat(df.a.values, df.b.str.len()),
                        'b':np.concatenate(df.b.values)})
Out[124]:
   a  b
0  1  1
1  1  2
2  2  2
3  2  3
4  2  4
5  3  5

OLD answer:

Try this:

In [89]: df.set_index('a', append=True).b.apply(pd.Series).stack().reset_index(level=[0, 2], drop=True).reset_index()
Out[89]:
   a    0
0  1  1.0
1  1  2.0
2  2  2.0
3  2  3.0
4  2  4.0
5  3  5.0

Or bit nicer solution provided by @Boud:

In [110]: df.set_index('a').b.apply(pd.Series).stack().reset_index(level=-1, drop=True).astype(int).reset_index()
Out[110]:
   a  0
0  1  1
1  1  2
2  2  2
3  2  3
4  2  4
5  3  5

You can use str.len for get length of lists which are repeated by numpy.repeat with flattening lists:

from  itertools import chain
import numpy as np

df2 = pd.DataFrame({
        0: np.repeat(df.iloc[:,0].values, df.iloc[:,1].str.len()),
        1: list(chain.from_iterable(df.iloc[:,1])),
        2: list(chain.from_iterable(df.iloc[:,2]))})

print (df2)        
   0  1  2
0  1  1  a
1  1  2  b
2  1  3  c
3  2  4  d
4  2  5  e
5  2  6  f
6  3  7  g
7  3  8  h

import itertools
import pandas as pd
import numpy as np

def melt_series(s):
    lengths = s.str.len().values
    flat = [i for i in itertools.chain.from_iterable(s.values.tolist())]
    idx = np.repeat(s.index.values, lengths)
    return pd.Series(flat, idx, name=s.name)


melt_series(df.type).to_frame().join(df.drop('type', 1)).reindex_axis(df.columns, 1)

setup

df = pd.DataFrame({'key':[1,2,3],
                   'type':[[1,3],[1,2,3],[1,2]],
                   'value':[5,1,8]})
df

Instead of this:

for i in range(len(trans_dates)):
    df[('T' + str(i + 1) + ' - Date')] = trans_dates[i]

Try this:

for i in range(len(trans_dates)):
    df.loc[i, ('T' + str(i + 1) + ' - Date')] = trans_dates[i]  

There are probably better ways to do this... df.merge or df.replace come to mind... it would be helpful if you posted what the input dataframe looked like and what the expected result is.

Did you try like this way to make it comma separated without header using to_csv?

df.to_csv('df.txt', index=False, header=None)

Without modifying too much of your code, I was able to use the enumerate() function instead of your i loop. I am using Python 3.

import pandas as pd 
import numpy as np 
# create sample dataframe
abc = ['a','b','c','d','e', 'f']
data = {'Text':['foo', 'bar', 'spam', 'eggs', 'ham', 'asdf'],
        'Number':[2, 2, 5, 2, 2, 3]}
df = pd.DataFrame.from_dict(data) 
df = df[['Text', 'Number']] # reordering df
print(df)

# Original question code:
# i=0
# for row1 in range(0,len(df)):
#     if df.iloc[row1]['Number']==2:
#         df.set_value(df.index[row1], 'Text',abc[i])
#         i=i+1

# Pseudocode: 
# if row value for the column 'number' is 2:
#     set the value for the column 'text' at an index to the value for abc[index]
for idx, row1 in enumerate(range(0, len(df))):
    if df.iloc[row1]['Number'] == 2:
        df.set_value(df.index[row1], 'Text', abc[idx])
print(df)

Please note that set_value is depreciated and will be removed in a future release.

Results:

    text  number
0     a       2
1     b       2
2  spam       5
3     d       2
4     e       2
5  asdf       3

FutureWarning: set_value is deprecated and will be removed in a future release. Please use .at[] or .iat[] accessors instead

You're over complicating this simple problem:

import csv
with open('test.txt') as f, open('out.csv','w') as f2:
    writer = csv.writer(f2, delimiter = '\n')
    for line in f:
        x, y = line.split() 
        writer.writerow([y]*int(x))

output:

>>> cat out.csv
3
7
7
7
7
8
8

You can use pandas for this:

import pandas as pd
import io

data = '''ID,dep_id
EMP1,"1,2,3"
EMP2,"4,5,6"'''

df = pd.read_csv(io.StringIO(data)) # insert your csv here
df['dep_id'] = ('['+df['dep_id']+ ']').apply(pd.eval) # turn string to list
df.explode('dep_id')df.explode('dep_id').reset_index(drop=True)

Output:

You can use concat and str.get_dummies:

print pd.concat([df['Id'], df['Field'].str.get_dummies(sep=",")], axis=1)
   Id  Economics  English-107  English-2  History  Java-2  Literature  \
0  56          1            1          0        1       0           0   
1  11          0            0          0        0       1           0   
2   6          0            0          1        0       0           1   
3  43          0            0          0        1       0           1   
4  14          0            1          0        0       1           0   

   Management  Mathematics  Philosophy  Web-development  
0           1            0           1                0  
1           0            0           0                1  
2           0            0           0                0  
3           0            1           1                0  
4           1            0           0                0  

If you need count values, you can use pivot_table (I add one string Economics for testing):

df1 = df['Field'].str.split(',',expand=True).stack()
                                            .groupby(level=0)
                                            .value_counts()
                                            .reset_index()
df1.columns=['a','b','c']
print df1.pivot_table(index='a',columns='b',values='c').fillna(0)
b  Economics  English-107  English-2  History  Java-2  Literature  Management  \
a                                                                               
0          2            1          0        1       0           0           1   
1          0            0          0        0       1           0           0   
2          0            0          1        0       0           1           0   
3          0            0          0        1       0           1           0   
4          0            1          0        0       1           0           1   

b  Mathematics  Philosophy  Web-development  
a                                            
0            0           1                0  
1            0           0                1  
2            0           0                0  
3            1           1                0  
4            0           0                0