Code is written to support multiple arguments, but fails when only one argument is given. sys.argv[1] +...

My current code uses a variable "word", so word = sys.argv[1] + sys.argv[2], which works great when I have two arguments, but when I only have one argument then i get IndexError: list index out of range

I am writing a program that Defines words by pulling the info from dictionary .com. I am following very closely a video tutorial because I am just trying to learn how to get a basic dictionary to work, (long term goal being write a translation app that is independent of google translate api or anything like that). I have a code that works in that it pulls the definition when you type in the word, but not when you type in 2 words, examples will be below. SO I added an argument, which worked for 2 words but no longer works for one word.

import requests

from bs4 import BeautifulSoup as bs

import sys



url = "https://www.dictionary.com/browse/"



word = sys.argv[1] + sys.argv[2]



url+= word

r = requests.get(url)

soup = bs(r.content, "lxml")





try:

    pos = soup.findAll("span", {"class" : "luna-pos"})[0].text #luna-pos is an html tag

    definitions = soup.findAll("ol")

    meanings = definitions[0].findChildren("li", recursive=False)

    print(word + ": " + pos)

    for (i,meaning) in enumerate(meanings):

        print(str(i + 1), meaning.text)

except:

     print("Word not found")

Expected results:
Print word, part of speech, and definition.
I.e Define the word (being a single argument) "Proper"
Also Define "Self Control" (two word argument)

asked Jan 1 at 19:44

Jadon Murphy

295

just check how many args you have before trying to access the non existant [2] index? basic if condition.

– Paritosh Singh
Jan 1 at 19:47

@ParitoshSingh Not really the best solution. Then what will happen with 3 arguments? 4? 100?

– DeepSpace
Jan 1 at 19:49

aye thats true.

– Paritosh Singh
Jan 1 at 20:00

add a comment |

import requests

from bs4 import BeautifulSoup as bs

import sys



url = "https://www.dictionary.com/browse/"



word = sys.argv[1] + sys.argv[2]



url+= word

r = requests.get(url)

soup = bs(r.content, "lxml")





try:

    pos = soup.findAll("span", {"class" : "luna-pos"})[0].text #luna-pos is an html tag

    definitions = soup.findAll("ol")

    meanings = definitions[0].findChildren("li", recursive=False)

    print(word + ": " + pos)

    for (i,meaning) in enumerate(meanings):

        print(str(i + 1), meaning.text)

except:

     print("Word not found")

Expected results:
Print word, part of speech, and definition.
I.e Define the word (being a single argument) "Proper"
Also Define "Self Control" (two word argument)

asked Jan 1 at 19:44

Jadon Murphy

295

just check how many args you have before trying to access the non existant [2] index? basic if condition.

– Paritosh Singh
Jan 1 at 19:47

@ParitoshSingh Not really the best solution. Then what will happen with 3 arguments? 4? 100?

– DeepSpace
Jan 1 at 19:49

aye thats true.

– Paritosh Singh
Jan 1 at 20:00

add a comment |

import requests

from bs4 import BeautifulSoup as bs

import sys



url = "https://www.dictionary.com/browse/"



word = sys.argv[1] + sys.argv[2]



url+= word

r = requests.get(url)

soup = bs(r.content, "lxml")





try:

    pos = soup.findAll("span", {"class" : "luna-pos"})[0].text #luna-pos is an html tag

    definitions = soup.findAll("ol")

    meanings = definitions[0].findChildren("li", recursive=False)

    print(word + ": " + pos)

    for (i,meaning) in enumerate(meanings):

        print(str(i + 1), meaning.text)

except:

     print("Word not found")

Expected results:
Print word, part of speech, and definition.
I.e Define the word (being a single argument) "Proper"
Also Define "Self Control" (two word argument)

asked Jan 1 at 19:44

Jadon Murphy

295

import requests

from bs4 import BeautifulSoup as bs

import sys



url = "https://www.dictionary.com/browse/"



word = sys.argv[1] + sys.argv[2]



url+= word

r = requests.get(url)

soup = bs(r.content, "lxml")





try:

    pos = soup.findAll("span", {"class" : "luna-pos"})[0].text #luna-pos is an html tag

    definitions = soup.findAll("ol")

    meanings = definitions[0].findChildren("li", recursive=False)

    print(word + ": " + pos)

    for (i,meaning) in enumerate(meanings):

        print(str(i + 1), meaning.text)

except:

     print("Word not found")

Expected results:
Print word, part of speech, and definition.
I.e Define the word (being a single argument) "Proper"
Also Define "Self Control" (two word argument)

python web-scraping translators

asked Jan 1 at 19:44

Jadon Murphy

295

asked Jan 1 at 19:44

Jadon Murphy

295

asked Jan 1 at 19:44

Jadon Murphy

295

asked Jan 1 at 19:44

Jadon Murphy

295

asked Jan 1 at 19:44

Jadon Murphy

295

just check how many args you have before trying to access the non existant [2] index? basic if condition.

– Paritosh Singh
Jan 1 at 19:47

@ParitoshSingh Not really the best solution. Then what will happen with 3 arguments? 4? 100?

– DeepSpace
Jan 1 at 19:49

aye thats true.

– Paritosh Singh
Jan 1 at 20:00

add a comment |

just check how many args you have before trying to access the non existant [2] index? basic if condition.

– Paritosh Singh
Jan 1 at 19:47

@ParitoshSingh Not really the best solution. Then what will happen with 3 arguments? 4? 100?

– DeepSpace
Jan 1 at 19:49

aye thats true.

– Paritosh Singh
Jan 1 at 20:00

just check how many args you have before trying to access the non existant [2] index? basic if condition.

– Paritosh Singh
Jan 1 at 19:47

@ParitoshSingh Not really the best solution. Then what will happen with 3 arguments? 4? 100?

– DeepSpace
Jan 1 at 19:49

aye thats true.

– Paritosh Singh
Jan 1 at 20:00

add a comment |

2 Answers
2

active

oldest

votes

This is a good use-case for slicing. You want everything after the first argument in the list, so as a slice this translates to [1:]. So just use:

word = ''.join(sys.argv[1:])

For example:

>>> args = ['scriptname.py', 'the']

>>> ''.join(args[1:])

'the'

>>> args = ['scriptname.py', 'self', 'control']

>>> ''.join(args[1:])

'selfcontrol'

answered Jan 1 at 19:47

juanpa.arrivillaga

38.8k33975

Hey thanks this seems to be working

– Jadon Murphy
Jan 1 at 19:50

@JadonMurphy Consider accepting the answer if it solved your problem.

– Bitto Bennichan
Jan 1 at 19:50

1

@bitto I will be it just says I have to wait some minutes. Thanks

– Jadon Murphy
Jan 1 at 19:58

add a comment |

Slicing:

You can use list slicing to avoid the error:

word = ''.join(sys.argv[1:]  # joins 1st to last ..

word = ''.join(sys.argv[1:3]  # 1st and 2nd without error

List slicing works even if the slice index is larger then the amount of elements in the list.

Testing:

The other way to handle it would be to test first then append:

if len(sys.argv) == 2:

    word = sys.argv[1]

elif len(sys.argv) >= 3:

    word = sys.argv[1]+sys.argv[2]

else:

    word = "" # no input given

Error handling:

Try it and handle the error if it happens (see Ask forgiveness not permission below)

try:

    word = sys.argv[1]+sys.argv[2]

except IndexError:

    word = sys.argv[1]

Readup:

Understanding Python's slice notation

len() check

"Ask forgiveness not permission" - explain

try: except:

edited Jan 1 at 20:05

answered Jan 1 at 19:50

Patrick Artner

25.3k62444

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

This is a good use-case for slicing. You want everything after the first argument in the list, so as a slice this translates to [1:]. So just use:

word = ''.join(sys.argv[1:])

For example:

>>> args = ['scriptname.py', 'the']

>>> ''.join(args[1:])

'the'

>>> args = ['scriptname.py', 'self', 'control']

>>> ''.join(args[1:])

'selfcontrol'

answered Jan 1 at 19:47

juanpa.arrivillaga

38.8k33975

Hey thanks this seems to be working

– Jadon Murphy
Jan 1 at 19:50

@JadonMurphy Consider accepting the answer if it solved your problem.

– Bitto Bennichan
Jan 1 at 19:50

1

@bitto I will be it just says I have to wait some minutes. Thanks

– Jadon Murphy
Jan 1 at 19:58

add a comment |

This is a good use-case for slicing. You want everything after the first argument in the list, so as a slice this translates to [1:]. So just use:

word = ''.join(sys.argv[1:])

For example:

>>> args = ['scriptname.py', 'the']

>>> ''.join(args[1:])

'the'

>>> args = ['scriptname.py', 'self', 'control']

>>> ''.join(args[1:])

'selfcontrol'

answered Jan 1 at 19:47

juanpa.arrivillaga

38.8k33975

Hey thanks this seems to be working

– Jadon Murphy
Jan 1 at 19:50

@JadonMurphy Consider accepting the answer if it solved your problem.

– Bitto Bennichan
Jan 1 at 19:50

1

@bitto I will be it just says I have to wait some minutes. Thanks

– Jadon Murphy
Jan 1 at 19:58

add a comment |

This is a good use-case for slicing. You want everything after the first argument in the list, so as a slice this translates to [1:]. So just use:

word = ''.join(sys.argv[1:])

For example:

>>> args = ['scriptname.py', 'the']

>>> ''.join(args[1:])

'the'

>>> args = ['scriptname.py', 'self', 'control']

>>> ''.join(args[1:])

'selfcontrol'

answered Jan 1 at 19:47

juanpa.arrivillaga

38.8k33975

This is a good use-case for slicing. You want everything after the first argument in the list, so as a slice this translates to [1:]. So just use:

word = ''.join(sys.argv[1:])

For example:

>>> args = ['scriptname.py', 'the']

>>> ''.join(args[1:])

'the'

>>> args = ['scriptname.py', 'self', 'control']

>>> ''.join(args[1:])

'selfcontrol'

answered Jan 1 at 19:47

juanpa.arrivillaga

38.8k33975

answered Jan 1 at 19:47

juanpa.arrivillaga

38.8k33975

answered Jan 1 at 19:47

juanpa.arrivillaga

38.8k33975

answered Jan 1 at 19:47

juanpa.arrivillaga

38.8k33975

Hey thanks this seems to be working

– Jadon Murphy
Jan 1 at 19:50

@JadonMurphy Consider accepting the answer if it solved your problem.

– Bitto Bennichan
Jan 1 at 19:50

1

@bitto I will be it just says I have to wait some minutes. Thanks

– Jadon Murphy
Jan 1 at 19:58

add a comment |

Hey thanks this seems to be working

– Jadon Murphy
Jan 1 at 19:50

@JadonMurphy Consider accepting the answer if it solved your problem.

– Bitto Bennichan
Jan 1 at 19:50

1

@bitto I will be it just says I have to wait some minutes. Thanks

– Jadon Murphy
Jan 1 at 19:58

Hey thanks this seems to be working

– Jadon Murphy
Jan 1 at 19:50

@JadonMurphy Consider accepting the answer if it solved your problem.

– Bitto Bennichan
Jan 1 at 19:50

@bitto I will be it just says I have to wait some minutes. Thanks

– Jadon Murphy
Jan 1 at 19:58

add a comment |

Slicing:

You can use list slicing to avoid the error:

word = ''.join(sys.argv[1:]  # joins 1st to last ..

word = ''.join(sys.argv[1:3]  # 1st and 2nd without error

List slicing works even if the slice index is larger then the amount of elements in the list.

Testing:

The other way to handle it would be to test first then append:

if len(sys.argv) == 2:

    word = sys.argv[1]

elif len(sys.argv) >= 3:

    word = sys.argv[1]+sys.argv[2]

else:

    word = "" # no input given

Error handling:

Try it and handle the error if it happens (see Ask forgiveness not permission below)

try:

    word = sys.argv[1]+sys.argv[2]

except IndexError:

    word = sys.argv[1]

Readup:

Understanding Python's slice notation

len() check

"Ask forgiveness not permission" - explain

try: except:

edited Jan 1 at 20:05

answered Jan 1 at 19:50

Patrick Artner

25.3k62444

add a comment |

Slicing:

You can use list slicing to avoid the error:

word = ''.join(sys.argv[1:]  # joins 1st to last ..

word = ''.join(sys.argv[1:3]  # 1st and 2nd without error

List slicing works even if the slice index is larger then the amount of elements in the list.

Testing:

The other way to handle it would be to test first then append:

if len(sys.argv) == 2:

    word = sys.argv[1]

elif len(sys.argv) >= 3:

    word = sys.argv[1]+sys.argv[2]

else:

    word = "" # no input given

Error handling:

Try it and handle the error if it happens (see Ask forgiveness not permission below)

try:

    word = sys.argv[1]+sys.argv[2]

except IndexError:

    word = sys.argv[1]

Readup:

Understanding Python's slice notation

len() check

"Ask forgiveness not permission" - explain

try: except:

edited Jan 1 at 20:05

answered Jan 1 at 19:50

Patrick Artner

25.3k62444

add a comment |

Slicing:

You can use list slicing to avoid the error:

word = ''.join(sys.argv[1:]  # joins 1st to last ..

word = ''.join(sys.argv[1:3]  # 1st and 2nd without error

List slicing works even if the slice index is larger then the amount of elements in the list.

Testing:

The other way to handle it would be to test first then append:

if len(sys.argv) == 2:

    word = sys.argv[1]

elif len(sys.argv) >= 3:

    word = sys.argv[1]+sys.argv[2]

else:

    word = "" # no input given

Error handling:

Try it and handle the error if it happens (see Ask forgiveness not permission below)

try:

    word = sys.argv[1]+sys.argv[2]

except IndexError:

    word = sys.argv[1]

Readup:

Understanding Python's slice notation

len() check

"Ask forgiveness not permission" - explain

try: except:

edited Jan 1 at 20:05

answered Jan 1 at 19:50

Patrick Artner

25.3k62444

Slicing:

You can use list slicing to avoid the error:

word = ''.join(sys.argv[1:]  # joins 1st to last ..

word = ''.join(sys.argv[1:3]  # 1st and 2nd without error

List slicing works even if the slice index is larger then the amount of elements in the list.

Testing:

The other way to handle it would be to test first then append:

if len(sys.argv) == 2:

    word = sys.argv[1]

elif len(sys.argv) >= 3:

    word = sys.argv[1]+sys.argv[2]

else:

    word = "" # no input given

Error handling:

Try it and handle the error if it happens (see Ask forgiveness not permission below)

try:

    word = sys.argv[1]+sys.argv[2]

except IndexError:

    word = sys.argv[1]

Readup:

Understanding Python's slice notation

len() check

"Ask forgiveness not permission" - explain

try: except:

edited Jan 1 at 20:05

answered Jan 1 at 19:50

Patrick Artner

25.3k62444

edited Jan 1 at 20:05

answered Jan 1 at 19:50

Patrick Artner

25.3k62444

answered Jan 1 at 19:50

Patrick Artner

25.3k62444

answered Jan 1 at 19:50

Patrick Artner

25.3k62444

add a comment |

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