Counting lowercase and uppercase letters in a string in Python












4












$begingroup$


I am writing a program to count the number of uppercase and lowercase letters in a string. I came up with something that works, but as I am still a beginner I have a feeling writing the code this way is probably considered "clumsy."



Here is what I have: 



stri = input("Give me a phrase:")

stri_up = 0

stri_lo = 0

for i in stri:

    if i.isupper():

        stri_up += 1

    if i.islower():

        stri_lo += 1

print("The number of uppercase letters in your phrase is:", stri_up)

print("The number of lowercase letters in your phrase is:", stri_lo)


Output:



Give me a phrase: tHe Sun is sHininG

The number of uppercase letters in your phrase is: 4

The number of lowercase letters in your phrase is: 11


I would like to learn how to write neat, beautiful code so I am wondering if there is a more efficient and elegant way to code this.






python beginner python-3.x strings







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migrated from stackoverflow.com 6 hours ago


This question came from our site for professional and enthusiast programmers.


















  • $begingroup$
    @juanpa.arrivillaga All critiques belong in answers, not comments.
    $endgroup$
    – 200_success
    4 hours ago
















4












$begingroup$


I am writing a program to count the number of uppercase and lowercase letters in a string. I came up with something that works, but as I am still a beginner I have a feeling writing the code this way is probably considered "clumsy."



Here is what I have: 



stri = input("Give me a phrase:")

stri_up = 0

stri_lo = 0

for i in stri:

    if i.isupper():

        stri_up += 1

    if i.islower():

        stri_lo += 1

print("The number of uppercase letters in your phrase is:", stri_up)

print("The number of lowercase letters in your phrase is:", stri_lo)


Output:



Give me a phrase: tHe Sun is sHininG

The number of uppercase letters in your phrase is: 4

The number of lowercase letters in your phrase is: 11


I would like to learn how to write neat, beautiful code so I am wondering if there is a more efficient and elegant way to code this.






python beginner python-3.x strings







share|improve this question











$endgroup$



migrated from stackoverflow.com 6 hours ago


This question came from our site for professional and enthusiast programmers.


















  • $begingroup$
    @juanpa.arrivillaga All critiques belong in answers, not comments.
    $endgroup$
    – 200_success
    4 hours ago














4












4








4





$begingroup$


I am writing a program to count the number of uppercase and lowercase letters in a string. I came up with something that works, but as I am still a beginner I have a feeling writing the code this way is probably considered "clumsy."



Here is what I have: 



stri = input("Give me a phrase:")

stri_up = 0

stri_lo = 0

for i in stri:

    if i.isupper():

        stri_up += 1

    if i.islower():

        stri_lo += 1

print("The number of uppercase letters in your phrase is:", stri_up)

print("The number of lowercase letters in your phrase is:", stri_lo)


Output:



Give me a phrase: tHe Sun is sHininG

The number of uppercase letters in your phrase is: 4

The number of lowercase letters in your phrase is: 11


I would like to learn how to write neat, beautiful code so I am wondering if there is a more efficient and elegant way to code this.






python beginner python-3.x strings







share|improve this question











$endgroup$




I am writing a program to count the number of uppercase and lowercase letters in a string. I came up with something that works, but as I am still a beginner I have a feeling writing the code this way is probably considered "clumsy."



Here is what I have: 



stri = input("Give me a phrase:")

stri_up = 0

stri_lo = 0

for i in stri:

    if i.isupper():

        stri_up += 1

    if i.islower():

        stri_lo += 1

print("The number of uppercase letters in your phrase is:", stri_up)

print("The number of lowercase letters in your phrase is:", stri_lo)


Output:



Give me a phrase: tHe Sun is sHininG

The number of uppercase letters in your phrase is: 4

The number of lowercase letters in your phrase is: 11


I would like to learn how to write neat, beautiful code so I am wondering if there is a more efficient and elegant way to code this.






python beginner python-3.x strings




python beginner python-3.x strings






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 4 hours ago









200_success

129k15153415




129k15153415










asked 7 hours ago









Madrid_dataladyMadrid_datalady

211




211




migrated from stackoverflow.com 6 hours ago


This question came from our site for professional and enthusiast programmers.









migrated from stackoverflow.com 6 hours ago


This question came from our site for professional and enthusiast programmers.














  • $begingroup$
    @juanpa.arrivillaga All critiques belong in answers, not comments.
    $endgroup$
    – 200_success
    4 hours ago


















  • $begingroup$
    @juanpa.arrivillaga All critiques belong in answers, not comments.
    $endgroup$
    – 200_success
    4 hours ago
















$begingroup$
@juanpa.arrivillaga All critiques belong in answers, not comments.
$endgroup$
– 200_success
4 hours ago




$begingroup$
@juanpa.arrivillaga All critiques belong in answers, not comments.
$endgroup$
– 200_success
4 hours ago










1 Answer
1






active

oldest

votes


















5












$begingroup$

Your code is mostly fine. I'd suggest more meaningful names for variables, e.g. i is typically a name for integer/index variables; since you're iterating over letters/characters, you might choose c, char, let, or letter. For stri, you might just name it phrase (that's what you asked for from the user after all). You get the idea. Make the names self-documenting.



Arguably you could make it look "prettier" by performing a single pass per test, replacing:



stri_up = 0

stri_lo = 0

for i in stri:

    if i.isupper():

        stri_up += 1

    if i.islower():

        stri_lo += 1


with:



stri_up = sum(1 for let in stri if let.isupper())

stri_lo = sum(1 for let in stri if let.islower())


That's in theory less efficient, since it has to traverse stri twice, while your original code only does it once, but in practice it's likely faster; on the CPython reference interpreter, sum is highly optimized for this case and avoids constructing a bunch of intermediate int objects while summing.






share|improve this answer








New contributor




ShadowRanger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$









  • 2




    $begingroup$
    You can just do sum(c.isupper() for c in phrase), because boolean will be treated as 0 or 1 when summing.
    $endgroup$
    – 200_success
    4 hours ago






  • 2




    $begingroup$
    @200_success: True, but I'm using dirty knowledge here; the sum fast path only fires for int (PyLong_Object at C layer) exactly (no int subclasses accepted, including bool); yielding bool blocks that optimization (and involves a lot more yields from the genexpr that can be avoided). Plus, I consider it more obvious to actually sum integers conditionally; using bool for numeric value is perfectly legal, just a little more magical than necessary, given the minimal benefit.
    $endgroup$
    – ShadowRanger
    3 hours ago












  • $begingroup$
    Just for comparison, a microbenchmark where stri/phrase is just one of each ASCII character (''.join(map(chr, range(128)))), takes 15.3 µs to complete on my computer using your code, vs. 10.5 µs for summing hardcoded 1s conditionally.
    $endgroup$
    – ShadowRanger
    3 hours ago













Your Answer





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1 Answer
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1 Answer
1






active

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active

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active

oldest

votes









5












$begingroup$

Your code is mostly fine. I'd suggest more meaningful names for variables, e.g. i is typically a name for integer/index variables; since you're iterating over letters/characters, you might choose c, char, let, or letter. For stri, you might just name it phrase (that's what you asked for from the user after all). You get the idea. Make the names self-documenting.



Arguably you could make it look "prettier" by performing a single pass per test, replacing:



stri_up = 0

stri_lo = 0

for i in stri:

    if i.isupper():

        stri_up += 1

    if i.islower():

        stri_lo += 1


with:



stri_up = sum(1 for let in stri if let.isupper())

stri_lo = sum(1 for let in stri if let.islower())


That's in theory less efficient, since it has to traverse stri twice, while your original code only does it once, but in practice it's likely faster; on the CPython reference interpreter, sum is highly optimized for this case and avoids constructing a bunch of intermediate int objects while summing.






share|improve this answer








New contributor




ShadowRanger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$









  • 2




    $begingroup$
    You can just do sum(c.isupper() for c in phrase), because boolean will be treated as 0 or 1 when summing.
    $endgroup$
    – 200_success
    4 hours ago






  • 2




    $begingroup$
    @200_success: True, but I'm using dirty knowledge here; the sum fast path only fires for int (PyLong_Object at C layer) exactly (no int subclasses accepted, including bool); yielding bool blocks that optimization (and involves a lot more yields from the genexpr that can be avoided). Plus, I consider it more obvious to actually sum integers conditionally; using bool for numeric value is perfectly legal, just a little more magical than necessary, given the minimal benefit.
    $endgroup$
    – ShadowRanger
    3 hours ago












  • $begingroup$
    Just for comparison, a microbenchmark where stri/phrase is just one of each ASCII character (''.join(map(chr, range(128)))), takes 15.3 µs to complete on my computer using your code, vs. 10.5 µs for summing hardcoded 1s conditionally.
    $endgroup$
    – ShadowRanger
    3 hours ago


















5












$begingroup$

Your code is mostly fine. I'd suggest more meaningful names for variables, e.g. i is typically a name for integer/index variables; since you're iterating over letters/characters, you might choose c, char, let, or letter. For stri, you might just name it phrase (that's what you asked for from the user after all). You get the idea. Make the names self-documenting.



Arguably you could make it look "prettier" by performing a single pass per test, replacing:



stri_up = 0

stri_lo = 0

for i in stri:

    if i.isupper():

        stri_up += 1

    if i.islower():

        stri_lo += 1


with:



stri_up = sum(1 for let in stri if let.isupper())

stri_lo = sum(1 for let in stri if let.islower())


That's in theory less efficient, since it has to traverse stri twice, while your original code only does it once, but in practice it's likely faster; on the CPython reference interpreter, sum is highly optimized for this case and avoids constructing a bunch of intermediate int objects while summing.






share|improve this answer








New contributor




ShadowRanger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$









  • 2




    $begingroup$
    You can just do sum(c.isupper() for c in phrase), because boolean will be treated as 0 or 1 when summing.
    $endgroup$
    – 200_success
    4 hours ago






  • 2




    $begingroup$
    @200_success: True, but I'm using dirty knowledge here; the sum fast path only fires for int (PyLong_Object at C layer) exactly (no int subclasses accepted, including bool); yielding bool blocks that optimization (and involves a lot more yields from the genexpr that can be avoided). Plus, I consider it more obvious to actually sum integers conditionally; using bool for numeric value is perfectly legal, just a little more magical than necessary, given the minimal benefit.
    $endgroup$
    – ShadowRanger
    3 hours ago












  • $begingroup$
    Just for comparison, a microbenchmark where stri/phrase is just one of each ASCII character (''.join(map(chr, range(128)))), takes 15.3 µs to complete on my computer using your code, vs. 10.5 µs for summing hardcoded 1s conditionally.
    $endgroup$
    – ShadowRanger
    3 hours ago
















5












5








5





$begingroup$

Your code is mostly fine. I'd suggest more meaningful names for variables, e.g. i is typically a name for integer/index variables; since you're iterating over letters/characters, you might choose c, char, let, or letter. For stri, you might just name it phrase (that's what you asked for from the user after all). You get the idea. Make the names self-documenting.



Arguably you could make it look "prettier" by performing a single pass per test, replacing:



stri_up = 0

stri_lo = 0

for i in stri:

    if i.isupper():

        stri_up += 1

    if i.islower():

        stri_lo += 1


with:



stri_up = sum(1 for let in stri if let.isupper())

stri_lo = sum(1 for let in stri if let.islower())


That's in theory less efficient, since it has to traverse stri twice, while your original code only does it once, but in practice it's likely faster; on the CPython reference interpreter, sum is highly optimized for this case and avoids constructing a bunch of intermediate int objects while summing.






share|improve this answer








New contributor




ShadowRanger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$



Your code is mostly fine. I'd suggest more meaningful names for variables, e.g. i is typically a name for integer/index variables; since you're iterating over letters/characters, you might choose c, char, let, or letter. For stri, you might just name it phrase (that's what you asked for from the user after all). You get the idea. Make the names self-documenting.



Arguably you could make it look "prettier" by performing a single pass per test, replacing:



stri_up = 0

stri_lo = 0

for i in stri:

    if i.isupper():

        stri_up += 1

    if i.islower():

        stri_lo += 1


with:



stri_up = sum(1 for let in stri if let.isupper())

stri_lo = sum(1 for let in stri if let.islower())


That's in theory less efficient, since it has to traverse stri twice, while your original code only does it once, but in practice it's likely faster; on the CPython reference interpreter, sum is highly optimized for this case and avoids constructing a bunch of intermediate int objects while summing.







share|improve this answer








New contributor




ShadowRanger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this answer



share|improve this answer






New contributor




ShadowRanger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









answered 6 hours ago









ShadowRangerShadowRanger

1512




1512




New contributor




ShadowRanger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





ShadowRanger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






ShadowRanger is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 2




    $begingroup$
    You can just do sum(c.isupper() for c in phrase), because boolean will be treated as 0 or 1 when summing.
    $endgroup$
    – 200_success
    4 hours ago






  • 2




    $begingroup$
    @200_success: True, but I'm using dirty knowledge here; the sum fast path only fires for int (PyLong_Object at C layer) exactly (no int subclasses accepted, including bool); yielding bool blocks that optimization (and involves a lot more yields from the genexpr that can be avoided). Plus, I consider it more obvious to actually sum integers conditionally; using bool for numeric value is perfectly legal, just a little more magical than necessary, given the minimal benefit.
    $endgroup$
    – ShadowRanger
    3 hours ago












  • $begingroup$
    Just for comparison, a microbenchmark where stri/phrase is just one of each ASCII character (''.join(map(chr, range(128)))), takes 15.3 µs to complete on my computer using your code, vs. 10.5 µs for summing hardcoded 1s conditionally.
    $endgroup$
    – ShadowRanger
    3 hours ago
















  • 2




    $begingroup$
    You can just do sum(c.isupper() for c in phrase), because boolean will be treated as 0 or 1 when summing.
    $endgroup$
    – 200_success
    4 hours ago






  • 2




    $begingroup$
    @200_success: True, but I'm using dirty knowledge here; the sum fast path only fires for int (PyLong_Object at C layer) exactly (no int subclasses accepted, including bool); yielding bool blocks that optimization (and involves a lot more yields from the genexpr that can be avoided). Plus, I consider it more obvious to actually sum integers conditionally; using bool for numeric value is perfectly legal, just a little more magical than necessary, given the minimal benefit.
    $endgroup$
    – ShadowRanger
    3 hours ago












  • $begingroup$
    Just for comparison, a microbenchmark where stri/phrase is just one of each ASCII character (''.join(map(chr, range(128)))), takes 15.3 µs to complete on my computer using your code, vs. 10.5 µs for summing hardcoded 1s conditionally.
    $endgroup$
    – ShadowRanger
    3 hours ago










2




2




$begingroup$
You can just do sum(c.isupper() for c in phrase), because boolean will be treated as 0 or 1 when summing.
$endgroup$
– 200_success
4 hours ago




$begingroup$
You can just do sum(c.isupper() for c in phrase), because boolean will be treated as 0 or 1 when summing.
$endgroup$
– 200_success
4 hours ago




2




2




$begingroup$
@200_success: True, but I'm using dirty knowledge here; the sum fast path only fires for int (PyLong_Object at C layer) exactly (no int subclasses accepted, including bool); yielding bool blocks that optimization (and involves a lot more yields from the genexpr that can be avoided). Plus, I consider it more obvious to actually sum integers conditionally; using bool for numeric value is perfectly legal, just a little more magical than necessary, given the minimal benefit.
$endgroup$
– ShadowRanger
3 hours ago






$begingroup$
@200_success: True, but I'm using dirty knowledge here; the sum fast path only fires for int (PyLong_Object at C layer) exactly (no int subclasses accepted, including bool); yielding bool blocks that optimization (and involves a lot more yields from the genexpr that can be avoided). Plus, I consider it more obvious to actually sum integers conditionally; using bool for numeric value is perfectly legal, just a little more magical than necessary, given the minimal benefit.
$endgroup$
– ShadowRanger
3 hours ago














$begingroup$
Just for comparison, a microbenchmark where stri/phrase is just one of each ASCII character (''.join(map(chr, range(128)))), takes 15.3 µs to complete on my computer using your code, vs. 10.5 µs for summing hardcoded 1s conditionally.
$endgroup$
– ShadowRanger
3 hours ago






$begingroup$
Just for comparison, a microbenchmark where stri/phrase is just one of each ASCII character (''.join(map(chr, range(128)))), takes 15.3 µs to complete on my computer using your code, vs. 10.5 µs for summing hardcoded 1s conditionally.
$endgroup$
– ShadowRanger
3 hours ago




















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日野市

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body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} 日野市 日野市 日文轉寫  • 日文 日野市  • 平假名 ひのし  • 罗马字 Hino-shi 金剛寺（關東三十六不動尊靈場第9號）的不動堂 市旗 徽章 日野市在東京都的位置 日野市 日野市在日本的位置 坐标： 35°40′17″N 139°23′42″E  /  35.6714°N 139.395°E  / 35.6714; 139.395 国家   日本 地方 關東地方 都道府縣 東京都 接鄰行政區 府中市、國立市、立川市、昭島市、八王子市、多摩市 政府  • 市長 大坪冬彦 面积  •  总计 27.53 平方公里（10.63 平方英里） 人口 （2014年12月1日）  • 總計 183,323  • 密度 6,660/平方公里（17,200/平方英里） 象徵  • 市樹 樫  • 市花 菊花  • 市鳥 普通翠鳥 时区 日本標準時間 （UTC+9） 地方公共團體編號 13212-8 邮政编码 〒 191-8686 市役所地址 日野市神明一丁目12番1號 電話號碼 +81-42-585-1111 法人編號 1000020132128 網站 http://www.city.hino.lg.jp/ 人口：日野市官方網頁 日野市 （日语： 日野市 ／ ひのし   Hino shi   * / ? ）為一位于東京都（不含島嶼部分）中央地帶的城市。從東京站乘坐中央線特別快速列車45分鐘即可到該市。面積27.53km²。1963年（昭和38年）11月3日，該市開始實行市制，為全日本第559個市。汽車企業日野自動車的總部位於這裡。 往東京都特別區部的通勤率為20.9%，往八王子市的通勤率為12.0%（平成22年國勢調査）。 目录 1 概要 ...
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Tu-95轟炸機

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body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} 图-95 Ту–95 北约代号：熊（Bear） 一架在安格斯空军基地的图-95MS 概觀 類型 战略轰炸机 代號 北约代号： Bear （熊） 乘員 7名 駕駛員2名，機尾炮手1名，其他人員4名 首飛 1952年11月12日 服役 1956年 設計 圖波列夫設計局 產量 超过500架 現況 现役 主要用戶   蘇聯   俄羅斯 衍生機型 图-114、图-119、Tu-95 技术数据 長度 49.50米（162呎5吋） 翼展 51.10米（167呎8吋） 高度 12.12米（39呎9吋） 翼面積 310平方米（3,330平方呎） 空重 90,000公斤（198,000磅） 最大起飛重量 188,000公斤（414,500磅） 發動機 4具库兹涅佐夫NK-12MV型渦輪螺旋槳發動機 功率 4×11,000千瓦（14,800馬力） 性能數據 最大速度 925公里／時（500節，575哩／時） 爬升率 10米／秒（2,000呎／分） 最大升限 12,000米（39,000呎） 最大航程 15,000公里（8,100海浬，9,400哩） 翼負荷 606公斤／平方米（124磅／平方呎） 推重比 235 W／公斤（0.143馬力／磅） 武器装备 機炮 1具/2具AM-23型23公釐雷達控制機砲（機尾） 飛彈 空对地导弹： Kh-20、Kh-22 Kh-26、Kh-55 炸彈 60枚FAB-250炸彈，或 30枚FAB-500炸彈。 其他 载弹量（包括飞弹）：最多15,000公斤（33,000磅） Tu-95熊式MR型機 图-95 （俄语： Ту–95 ， 英语： Tu-95 ），北約代號： Bear （ 熊 ），是蘇聯圖波列夫設計局所研製，是全世界唯一服役的大型四渦輪螺旋槳發動機...
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