Quite newbie, I need help to search and modify some elements of a tuple (Python 2.7).
For example, how to replace "aaa" in first element into "bbb" ?
list=[('aaastring',245878L,2475L,'anotherstring', ...),
('bbbstring',7894578L,456897L,'yetanother',...),
('aaaform',2445L,325478L,'dummy',...),
(...)
]
# you should avoid names which are used for
# bult-in function/classes
# list is one of them
data = [('aaastring', 245878L, 2475L, 'anotherstring'),
('bbbstring', 7894578L, 456897L, 'yetanother'),
('aaaform', 2445L, 325478L, 'dummy')]
new_list = []
for row in data:
# row is a tuple
# a tuple is imuatble
# you need to create a new one
# or exchange the tuple with a list
row = list(row)
row[0] = 'Replacement for col0 of all rows'
new_list.append(row)Content of new_list
Output:
[['Replacement for col0 of all rows', 245878L, 2475L, 'anotherstring'],
['Replacement for col0 of all rows', 7894578L, 456897L, 'yetanother'],
['Replacement for col0 of all rows', 2445L, 325478L, 'dummy']]
Instead of overwriting the first column blindly, you can pass them into a function, which does the work for manipulation based on original data of the object in the first column.
By the way, use Python 3. It gives you super powers.
(Aug-11-2017, 06:27 AM)DeaD_EyE Wrote: [ -> ]# you should avoid names which are used for
# bult-in function/classes
# list is one of them
data = [('aaastring', 245878L, 2475L, 'anotherstring'),
('bbbstring', 7894578L, 456897L, 'yetanother'),
('aaaform', 2445L, 325478L, 'dummy')]
new_list = []
for row in data:
# row is a tuple
# a tuple is imuatble
# you need to create a new one
# or exchange the tuple with a list
row = list(row)
row[0] = 'Replacement for col0 of all rows'
new_list.append(row)Content of new_list
Output:
[['Replacement for col0 of all rows', 245878L, 2475L, 'anotherstring'],
['Replacement for col0 of all rows', 7894578L, 456897L, 'yetanother'],
['Replacement for col0 of all rows', 2445L, 325478L, 'dummy']]
Instead of overwriting the first column blindly, you can pass them into a function, which does the work for manipulation based on original data of the object in the first column.
By the way, use Python 3. It gives you super powers.
Thanks, work great !
However, new-list is not the same as data ?
data is of the form [(...),(...)]
new-list is of the form [[...],[...]]
Do I miss something ?
(Aug-12-2017, 06:43 AM)webdef Wrote: [ -> ] (Aug-11-2017, 06:27 AM)DeaD_EyE Wrote: [ -> ]# you should avoid names which are used for
# bult-in function/classes
# list is one of them
data = [('aaastring', 245878L, 2475L, 'anotherstring'),
('bbbstring', 7894578L, 456897L, 'yetanother'),
('aaaform', 2445L, 325478L, 'dummy')]
new_list = []
for row in data:
# row is a tuple
# a tuple is imuatble
# you need to create a new one
# or exchange the tuple with a list
row = list(row)
row[0] = 'Replacement for col0 of all rows'
new_list.append(row)Content of new_list
Output:
[['Replacement for col0 of all rows', 245878L, 2475L, 'anotherstring'],
['Replacement for col0 of all rows', 7894578L, 456897L, 'yetanother'],
['Replacement for col0 of all rows', 2445L, 325478L, 'dummy']]
Instead of overwriting the first column blindly, you can pass them into a function, which does the work for manipulation based on original data of the object in the first column.
By the way, use Python 3. It gives you super powers.
Thanks, work great !
However, new-list is not the same as data ?
data is of the form [(...),(...)]
new-list is of the form [[...],[...]]
Do I miss something ?
Solution could be something like:
row=list(row)
mywork = row[0]
mywork = mywork.replace("aaa","bbb",1)
row[0] = mywork
row2=tuple(row)
new_list.append(row2)
Is this correct for Python ?
That would work, but you can modify the first item in the row directly:
row[0].replace('aaa', 'bbb', 1)You can also tuple and append at the same time:
new_list.append(tuple(row))