python dictionary rotation result
i have this dictionary:
dict1 = {'Name': ['Mary','Bob'],'Surname': ['White','Red']}
if I do the print I have this output:
Surname White
Red
Name Mary
Bob
but I would like this output
Name Surname
Mary White
Bob Red
how can I do it without using pandas?
python
asked Nov 22 '18 at 18:30
luana nastasiluana nastasi
113
add a comment |
i have this dictionary:
dict1 = {'Name': ['Mary','Bob'],'Surname': ['White','Red']}
if I do the print I have this output:
Surname White
Red
Name Mary
Bob
but I would like this output
Name Surname
Mary White
Bob Red
how can I do it without using pandas?
python
asked Nov 22 '18 at 18:30
luana nastasiluana nastasi
113
add a comment |
i have this dictionary:
dict1 = {'Name': ['Mary','Bob'],'Surname': ['White','Red']}
if I do the print I have this output:
Surname White
Red
Name Mary
Bob
but I would like this output
Name Surname
Mary White
Bob Red
how can I do it without using pandas?
python
asked Nov 22 '18 at 18:30
luana nastasiluana nastasi
113
i have this dictionary:
dict1 = {'Name': ['Mary','Bob'],'Surname': ['White','Red']}
if I do the print I have this output:
Surname White
Red
Name Mary
Bob
but I would like this output
Name Surname
Mary White
Bob Red
how can I do it without using pandas?
python
python
asked Nov 22 '18 at 18:30
luana nastasiluana nastasi
113
asked Nov 22 '18 at 18:30
luana nastasiluana nastasi
113
asked Nov 22 '18 at 18:30
luana nastasiluana nastasi
113
asked Nov 22 '18 at 18:30
luana nastasiluana nastasi
113
asked Nov 22 '18 at 18:30
luana nastasiluana nastasi
113
113
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
Here's code that will sort keys of your dict (so you will get same out every time for same data) and print your dict in format you supplied with any amounts of keys and values of your dict.
dict1 = {'Name': ['Mary','Bob'],'Surname': ['White','Red']}
sorted_keys = sorted(dict1.keys())
spacing = 10
print_sheme = ('{:<' + str(spacing) + '}')*len(list(dict1.values())[0])
print(print_sheme.format(*sorted_keys))
for i in range(len(list(dict1.values())[0])):
values_to_print = [dict1[key][i] for key in sorted_keys]
print(print_sheme.format(*values_to_print))
Output:
Name Surname
Mary White
Bob Red
answered Nov 22 '18 at 18:46
Filip MłynarskiFilip Młynarski
1,7961413
add a comment |
A less general, but much shorter answer. Just iterate over both lists (but you can extend this to as how many lists you need to reference simultaneously):
print("{:10} {:10}".format("Name", "Surname"))
for (name, surname) in zip(dict1.get('Name'), dict1.get('Surname')):
print("{:10} {:10}".format(name, surname))
answered Nov 22 '18 at 20:08
planetmakerplanetmaker
4,68421729
add a comment |
Here's a pretty straightforward implementation that handles arbitrary keys, uneven lists, and dynamically-sized columns:
def column_print(d, spacing=2):
columns = [[x] + y for x, y in zip(d.keys(), d.values())]
col_widths = [spacing + max([len(x) for x in e]) for e in columns]
for i in range(max([len(x) for x in columns])):
for w, col in zip(col_widths, columns):
print((col[i] if i < len(col) else "").ljust(w), end="")
print()
column_print({'Name': ['Mary','Bob','Mark'], 'Surname': ['White','Red','Blue','Black'], 'MI': ['S','A']})
Output:
Name Surname MI
Mary White S
Bob Red A
Mark Blue
Black
Try it!
answered Nov 22 '18 at 18:55
ggorlenggorlen
7,1883826
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Here's code that will sort keys of your dict (so you will get same out every time for same data) and print your dict in format you supplied with any amounts of keys and values of your dict.
dict1 = {'Name': ['Mary','Bob'],'Surname': ['White','Red']}
sorted_keys = sorted(dict1.keys())
spacing = 10
print_sheme = ('{:<' + str(spacing) + '}')*len(list(dict1.values())[0])
print(print_sheme.format(*sorted_keys))
for i in range(len(list(dict1.values())[0])):
values_to_print = [dict1[key][i] for key in sorted_keys]
print(print_sheme.format(*values_to_print))
Output:
Name Surname
Mary White
Bob Red
answered Nov 22 '18 at 18:46
Filip MłynarskiFilip Młynarski
1,7961413
add a comment |
Here's code that will sort keys of your dict (so you will get same out every time for same data) and print your dict in format you supplied with any amounts of keys and values of your dict.
dict1 = {'Name': ['Mary','Bob'],'Surname': ['White','Red']}
sorted_keys = sorted(dict1.keys())
spacing = 10
print_sheme = ('{:<' + str(spacing) + '}')*len(list(dict1.values())[0])
print(print_sheme.format(*sorted_keys))
for i in range(len(list(dict1.values())[0])):
values_to_print = [dict1[key][i] for key in sorted_keys]
print(print_sheme.format(*values_to_print))
Output:
Name Surname
Mary White
Bob Red
answered Nov 22 '18 at 18:46
Filip MłynarskiFilip Młynarski
1,7961413
add a comment |
Here's code that will sort keys of your dict (so you will get same out every time for same data) and print your dict in format you supplied with any amounts of keys and values of your dict.
dict1 = {'Name': ['Mary','Bob'],'Surname': ['White','Red']}
sorted_keys = sorted(dict1.keys())
spacing = 10
print_sheme = ('{:<' + str(spacing) + '}')*len(list(dict1.values())[0])
print(print_sheme.format(*sorted_keys))
for i in range(len(list(dict1.values())[0])):
values_to_print = [dict1[key][i] for key in sorted_keys]
print(print_sheme.format(*values_to_print))
Output:
Name Surname
Mary White
Bob Red
answered Nov 22 '18 at 18:46
Filip MłynarskiFilip Młynarski
1,7961413
Here's code that will sort keys of your dict (so you will get same out every time for same data) and print your dict in format you supplied with any amounts of keys and values of your dict.
dict1 = {'Name': ['Mary','Bob'],'Surname': ['White','Red']}
sorted_keys = sorted(dict1.keys())
spacing = 10
print_sheme = ('{:<' + str(spacing) + '}')*len(list(dict1.values())[0])
print(print_sheme.format(*sorted_keys))
for i in range(len(list(dict1.values())[0])):
values_to_print = [dict1[key][i] for key in sorted_keys]
print(print_sheme.format(*values_to_print))
Output:
Name Surname
Mary White
Bob Red
answered Nov 22 '18 at 18:46
Filip MłynarskiFilip Młynarski
1,7961413
answered Nov 22 '18 at 18:46
Filip MłynarskiFilip Młynarski
1,7961413
answered Nov 22 '18 at 18:46
Filip MłynarskiFilip Młynarski
1,7961413
answered Nov 22 '18 at 18:46
Filip MłynarskiFilip Młynarski
1,7961413
1,7961413
add a comment |
add a comment |
A less general, but much shorter answer. Just iterate over both lists (but you can extend this to as how many lists you need to reference simultaneously):
print("{:10} {:10}".format("Name", "Surname"))
for (name, surname) in zip(dict1.get('Name'), dict1.get('Surname')):
print("{:10} {:10}".format(name, surname))
answered Nov 22 '18 at 20:08
planetmakerplanetmaker
4,68421729
add a comment |
A less general, but much shorter answer. Just iterate over both lists (but you can extend this to as how many lists you need to reference simultaneously):
print("{:10} {:10}".format("Name", "Surname"))
for (name, surname) in zip(dict1.get('Name'), dict1.get('Surname')):
print("{:10} {:10}".format(name, surname))
answered Nov 22 '18 at 20:08
planetmakerplanetmaker
4,68421729
add a comment |
A less general, but much shorter answer. Just iterate over both lists (but you can extend this to as how many lists you need to reference simultaneously):
print("{:10} {:10}".format("Name", "Surname"))
for (name, surname) in zip(dict1.get('Name'), dict1.get('Surname')):
print("{:10} {:10}".format(name, surname))
answered Nov 22 '18 at 20:08
planetmakerplanetmaker
4,68421729
A less general, but much shorter answer. Just iterate over both lists (but you can extend this to as how many lists you need to reference simultaneously):
print("{:10} {:10}".format("Name", "Surname"))
for (name, surname) in zip(dict1.get('Name'), dict1.get('Surname')):
print("{:10} {:10}".format(name, surname))
answered Nov 22 '18 at 20:08
planetmakerplanetmaker
4,68421729
edited Nov 22 '18 at 20:18
answered Nov 22 '18 at 20:08
planetmakerplanetmaker
4,68421729
answered Nov 22 '18 at 20:08
planetmakerplanetmaker
4,68421729
answered Nov 22 '18 at 20:08
planetmakerplanetmaker
4,68421729
4,68421729
add a comment |
add a comment |
Here's a pretty straightforward implementation that handles arbitrary keys, uneven lists, and dynamically-sized columns:
def column_print(d, spacing=2):
columns = [[x] + y for x, y in zip(d.keys(), d.values())]
col_widths = [spacing + max([len(x) for x in e]) for e in columns]
for i in range(max([len(x) for x in columns])):
for w, col in zip(col_widths, columns):
print((col[i] if i < len(col) else "").ljust(w), end="")
print()
column_print({'Name': ['Mary','Bob','Mark'], 'Surname': ['White','Red','Blue','Black'], 'MI': ['S','A']})
Output:
Name Surname MI
Mary White S
Bob Red A
Mark Blue
Black
Try it!
answered Nov 22 '18 at 18:55
ggorlenggorlen
7,1883826
add a comment |
Here's a pretty straightforward implementation that handles arbitrary keys, uneven lists, and dynamically-sized columns:
def column_print(d, spacing=2):
columns = [[x] + y for x, y in zip(d.keys(), d.values())]
col_widths = [spacing + max([len(x) for x in e]) for e in columns]
for i in range(max([len(x) for x in columns])):
for w, col in zip(col_widths, columns):
print((col[i] if i < len(col) else "").ljust(w), end="")
print()
column_print({'Name': ['Mary','Bob','Mark'], 'Surname': ['White','Red','Blue','Black'], 'MI': ['S','A']})
Output:
Name Surname MI
Mary White S
Bob Red A
Mark Blue
Black
Try it!
answered Nov 22 '18 at 18:55
ggorlenggorlen
7,1883826
add a comment |
Here's a pretty straightforward implementation that handles arbitrary keys, uneven lists, and dynamically-sized columns:
def column_print(d, spacing=2):
columns = [[x] + y for x, y in zip(d.keys(), d.values())]
col_widths = [spacing + max([len(x) for x in e]) for e in columns]
for i in range(max([len(x) for x in columns])):
for w, col in zip(col_widths, columns):
print((col[i] if i < len(col) else "").ljust(w), end="")
print()
column_print({'Name': ['Mary','Bob','Mark'], 'Surname': ['White','Red','Blue','Black'], 'MI': ['S','A']})
Output:
Name Surname MI
Mary White S
Bob Red A
Mark Blue
Black
Try it!
answered Nov 22 '18 at 18:55
ggorlenggorlen
7,1883826
Here's a pretty straightforward implementation that handles arbitrary keys, uneven lists, and dynamically-sized columns:
def column_print(d, spacing=2):
columns = [[x] + y for x, y in zip(d.keys(), d.values())]
col_widths = [spacing + max([len(x) for x in e]) for e in columns]
for i in range(max([len(x) for x in columns])):
for w, col in zip(col_widths, columns):
print((col[i] if i < len(col) else "").ljust(w), end="")
print()
column_print({'Name': ['Mary','Bob','Mark'], 'Surname': ['White','Red','Blue','Black'], 'MI': ['S','A']})
Output:
Name Surname MI
Mary White S
Bob Red A
Mark Blue
Black
Try it!
answered Nov 22 '18 at 18:55
ggorlenggorlen
7,1883826
edited Nov 22 '18 at 20:35
answered Nov 22 '18 at 18:55
ggorlenggorlen
7,1883826
answered Nov 22 '18 at 18:55
ggorlenggorlen
7,1883826
answered Nov 22 '18 at 18:55
ggorlenggorlen
7,1883826
7,1883826
add a comment |
add a comment |
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